 NCERT Solutions Class 10 Maths Chapter 2 Polynomials – Here are all the NCERT solutions for Class 10 Maths Chapter 2. This solution contains questions, answers, images, explanations of the complete Chapter 2 titled Polynomials of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 2 Polynomials. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 2 Polynomials in one place.

## NCERT Solutions Class 10 Maths Chapter 2 Polynomials

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 Class 10 Subject Maths Book Mathematics Chapter Number 2 Chapter Name Polynomials

### NCERT Solutions Class 10 Maths chapter 2 Polynomials

Class 10, Maths chapter 2, Polynomials solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Polynomials

Q.1: The graphs of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. Ans : (i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points. 
Q.1: Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer
(i) $$4 x^{2}-3 x+7$$
(ii) $$y^{2}+\sqrt{2}$$
(iii) $$3 \sqrt{t}+t\sqrt{2}$$
(iv) $$y+\frac{2}{y}$$
(v) $$x^{2}+y^{3}+t$$
Ans : (i) $$4 x^{2}-3 x+7$$
Yes, this expression is a polynomial in one variable x.
(ii) $$y^{2}+\sqrt{2}$$
Yes, this expression is a polynomial in one variable x.
(iii) $$3 \sqrt{t}+t \sqrt{2}$$
No It can be observed that the exponent of variable t in term $$3 \sqrt{t}_{\text { is }} \frac{1}{2}$$  which is not a whole number .Therefore this expression is not a polynomial.
(iv) $$y+\frac{2}{y}$$ No It can be observed that the exponent of variable t in term $$\frac{2}{y}_{\text { is }-1,}$$  which is not a whole number .Therefore this expression is not a polynomial.
(v) $$x^{2}+y^{3}+t$$ No It can be observed that this expression is a polynomial in 3 variables x,y and and t .Therefore , this expression is not a polynomial . 
Q.2: Write the coefficients of  $$x^{2}$$  in each of the following:
(i) $$2+x^{2}+x$$
(ii) $$2-x^{2}+x^{3}$$
(iii) $$\frac{\pi}{2} x^{2}+x$$
(iv) $$\sqrt{2} x-1$$
Ans : (i) $$2+x^{2}+x$$
In the above expression the coefficient of $$x^{2}$$ is $$1.$$
(ii) $$2-x^{2}+x^{3}$$
In the above expression the coefficient of $$x^{2}$$ is $$-1$$
(iii) $$\frac{\pi}{2} x^{2}+x$$
In the above expression the coefficient of $$x^{2}$$ is $$\frac{\pi}{2}$$
(iv) $$\sqrt{2} x-1$$  $$0 x^{2}+\sqrt{2} x-1$$ or $$0 x^{2}+\sqrt{2} x-1$$
In the above expression the coefficient of $$x^{2}$$ is $$0$$ 
Q.3: Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Ans : Degree of a polynomial is the highest power of the variable in the polynomial.
Binomial has two terms in it. Therefore, binomial of degree 35 can be written as $$x^{35}+x^{34}$$
Monomial has only one term in it. Therefore, monomial of degree 100 can be written as $$X^{100}$$ 
Q.4: Write the degree of each of the following polynomials
(i) $$5 x^{3}+4 x^{2}+7 x$$
(ii) $$4-y^{2}$$
(iii) $$5 t-\sqrt{7}$$
(iv) 3
Ans : Degree of a polynomial is the highest power of the variable in the polynomial.
(i) $$5 x^{3}+4 x^{2}+7 x$$
This is a polynomial in a variable x and the highest power of variable x is 3 . Therefore, the degree of this polynomial is 3.
(ii) $$4-y^{2}$$ This is a polynomial in variable y and the highest power of variable y is 2. Therefore, the degree of this polynomial is 2 .
(iii) $$5 t-\sqrt{7}$$ This is a polynomial in variable t and the highest power of variable t is 1. Therefore, the degree of this polynomial is 1.
(iv) 3 This is a constant polynomial. Degree of a constant polynomial is always 0. 
Q.5: Classify the following as linear, quadratic and cubic polynomials:
(i) $$x^{2}+x$$
(ii) $$x-x^{3}$$
(iii) $$y+y^{2}+4$$
(iv) $$1+x$$
(v) $$3 \mathrm{t}$$
(vi) $$r^{2}$$
(vii) $$7 x^{3}$$
Ans : Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively.

(i) $$x^{2}+x$$ is a quadratic polynomial as its degree is 2 .
(ii) $$x-x^{3}$$ is a cubic polynomial as its degree is 3 .
(iii) $$y+y^{2}+4$$ is a quadratic polynomial as its degree is 1.
(iv) $$1+x$$ is a linear polynomial as its degree is 1.
(v) $$3 t$$ is a linear polynomial as its degree is 1.
(vi) $$r^{2}$$ is a quadratic polynomial as its degree is 2.
(vii) $$7 x^{3}$$  s a cubic polynomial as its degree is 3 . 
Q.1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i)   $$\ x^{2}-2 x-8$$
(ii)  $$\ 4 s^{2}-4 s+1$$
(iii) $$\ 6 x^{2}-3-7x$$
(iv)  $$\ 4 u^{2}+8 u$$
(v)   $$\ t^{2}-15$$
(vi)  $$\ 3 x^{2}-x-4$$
Ans : The value of  $$\ x^{2}-2 x-8$$ is zero if x +2 = 0 or x —4 0
$$\Rightarrow x=-2 \text { or } x=4$$
Therefore, the zeroes of  $$\ x^{2}-2 x-8\ are —2 and 4. Now ( \ =-2+4=2=\frac{-(-2)}{1}=\frac{-(\text { Cofficient of } x)}{\text { Cofficient of } x^{2}}$$
$$\ =(-2) \times 4=-8=\frac{-8}{1}=\frac{\text { Constant term }}{\text { Cofficient of } x^{2}}$$

(ii)
$$4 s^{2}-4 s+1=(2 s-1)^{2}$$
$$\begin{array}{l}{\text { The value of } 4 s^{2}-4 s+1 \text { is zero when } 2 s-1=0, \text { i.e., }^{s=\frac{1}{2}}} \\ {\text { Therefore, the zeroes of } 4 s^{2}-4 s+1 \text { are } \frac{1}{2} \text { and } \frac{1}{2}}\end{array}$$
Sum of zeroes =$$\frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}=\frac{-(\text { Coefficient of } s)}{\left(\text { Coefficient of } s^{2}\right)}$$
Product of zeroes$$=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}=\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}$$

(iii)
$$6 x^{2}-3-7 x=6 x^{2}-7 x-3=(3 x+1)(2 x-3)$$
$$\begin{array}{l}{\text { The value of } 6 x^{2}-3-7 x \text { is zero when } 3 x+1=0 \text { or } 2 x-3=0, \text { i.e., }} \\ {x=\frac{-1}{3} \text { or } x=\frac{3}{2}} \\ {\text { Therefore, the zeroes of } 6 x^{2}-3-7 x \text { are } \frac{3}{3} \text { and } \frac{3}{2}}\end{array}$$

$$\begin{array}{l}{\text { Sum of zeroes = } \frac{-1}{3}+\frac{3}{2}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-(\text { Coefficient of } x)}{6}} \\ {\text { Product of zeroes }=\frac{-1}{3} \times \frac{3}{2}=\frac{-1}{2}=\frac{-3}{6}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}}\end{array}$$

(iv)
$$\begin{array}{l}{4 u^{2}+8 u=4 u^{2}+8 u+0} \\ {=4 u(u+2)}\end{array}$$
$$\begin{array}{l}{\text {The value of } 4 u^{2}+8 u \text { is zero when } 4 u=0 \text { or } u+2=0, \text { i.e., } u=0 \text { or }} \\ {u=-2} \\ {\text {Therefore, the zeroes of } 4 u^{2}+8 u \text { are } 0 \text { and }-2 \text { . }}\end{array}$$
Sum of zeroes$$=^{0+(-2)=-2=\frac{-(8)}{4}}=\frac{-(\text { Coefficient of } u)}{\text { Coefficient of } u^{2}}$$
Product of zeroes$$=^{0 \times(-2)}=0=\frac{0}{4}=\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}$$

(v)
$$\begin{array}{l}{t^{2}-15} \\ {=t^{2}-0.1-15} \\ {=(t-\sqrt{15})(t+\sqrt{15})}\end{array}$$
$$\begin{array}{l}{\text { The value of } t^{2}-15 \text { is zero when } t-\sqrt{15}=0 \text { or } t+\sqrt{15}=0, \text { l.e., when }} \\ {t=\sqrt{15} \text { or } t=-\sqrt{15}} \\ {\text { Therefore, the zeroes of } t^{2}-15 \text { are } \sqrt{15} \text { and - } \sqrt{15} \text { . }}\end{array}$$
Sum of zeroes =$$\sqrt{15}+(-\sqrt{15})=0=\frac{-0}{1}=\frac{-(\text { Coefficient of } t)}{\left(\text { Coefficient of } t^{2}\right)}$$
Product of zeroes=$$\left( \begin{array}{l}{\sqrt{15} )(-\sqrt{15})=-15=\frac{-15}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}}\end{array}\right.$$
$$\begin{array}{l}{\text { (vi) } 3 x^{2}-x-4} \\ {=(3 x-4)(x+1)} \\ {\text { The value of } 3 x^{2}-x-4 \text { is zero when } 3 x-4=0 \text { or } x+1=0, \text { i.e., }} \\ {\text { when } x=\frac{4}{3} \text { or } x=-1}\end{array}$$
$$\begin{array}{l}{\text { Therefore, the zeroes of } 3 x^{2}-x-4 \text { are }^{\frac{4}{3}} \text { and }-1} \\ {\text { Sum of zeroes }=\frac{4}{3}+(-1)=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}}\end{array}$$
Product of zeroes$$=\frac{4}{3}(-1)=\frac{-4}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$$ 
Q.2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $$\frac{1}{4},-1$$
(ii) $$\sqrt{2}, \frac{1}{3}$$
(iii) $$0, \sqrt{5}$$
(iv) 1,1
(v) $$-\frac{1}{4}, \frac{1}{4}$$
(vi) $$4,1$$
Ans : (i) Let and ß are the zeroes of the polynomial $$a x^{2}+b x+c$$, then we have

$$\alpha+\beta=\frac{1}{4}=\frac{-b}{a}$$
On comparing, $$\alpha \beta=-1=\frac{-4}{4}=\frac{C}{a}$$
a=4,b=-1,and c=-4

(ii) Let and ß are the zeroes of the polynomial $$a x^{2}+b x+c$$, then we have
$$\alpha+\beta=\sqrt{2}=\frac{3 \sqrt{2}}{3}=\frac{-b}{a}$$
On comparing,
$$\alpha \beta=\frac{1}{3}=\frac{c}{a}$$
$$a=3, b=-3 \sqrt{2} \text { and } c=1$$
Hence, the required quadratic polynomial is $$x^{2}+0 . x+\sqrt{5}$$

(iii) Let a and are the zeroes or the polynomial $$a x^{2}+b x+c$$. then we have
$$\alpha+\beta=0=\frac{0}{1}=\frac{-b}{a}$$
on comparing  $$\alpha \beta=\sqrt{5}=\frac{\sqrt{5}}{1}=\frac{c}{a}$$
Hence, the required quadratic polynomial is : $$x^{2}+0 . x+\sqrt{5}$$

(iv) Let a and are the zeroes Of the polynomial $$a x^{2}+b x+c$$, then we have
$$\alpha+\beta=1=\frac{1}{1}=\frac{-b}{a}$$
$$\alpha \beta=1=\frac{1}{1}=\frac{c}{a}$$
on comparing, $$a=1, \quad b=-1 \text { and } c=1$$
Hence. the required quadratic polynomial is $$x^{2}-x+1$$

(v) Let a and are the zeroes Of the polynomial $$a x^{2}+b x+c$$, then we have
$$\alpha+\beta=1=\frac{-1}{4}=\frac{-b}{a}$$
$$\alpha \beta=1=\frac{1}{4}=\frac{c}{a}$$
on comparing,
a=4,b=1,and c=1
Hence. the required quadratic polynomial is $$4 x^{2}+x+1$$

(vi) Let a and are the zeroes Ofthe polynomial axa + bx + c, then we have
$$\alpha+\beta=4=\frac{4}{1}=\frac{-b}{a}$$
$$\alpha \beta=1=\frac{1}{1}=\frac{c}{a}$$
on comparing ,
a=1, b=-4 and c=1
Hence, the required quadratic polynomial is $$x^{2}-4 x+1$$ 
Q.1: Find the value of the polynomial $$5 x-4 x^{2}+3$$ at

(i)  x=0
(ii)  x=-1
(iii)  x=2
Ans : (i) $$p(x)=5 x-4 x^{2}+3$$
\begin{aligned} p(0) &=5(0)-4(0)^{2}+3 \\ &=3 \end{aligned}
(ii) $$p(x)=5 x-4 x^{2}+3$$
\begin{aligned} p(-1) &=5(-1)-4(-1)^{2}+3 \\ &=-5-4(1)+3=-6 \end{aligned}

(iii) $$p(x)=5 x-4 x^{2}+3$$
\begin{aligned} p(2) &=5(2)-4(2)^{2}+3 \\ &=10-16+3=-3 \end{aligned} 
Q.2: Find  p(0), p(1) and  p(2)  for each of the following polynomials;

(i) $$p(y)=y^{2}-y+1$$
(ii) $$p(t)=2+t+2 t^{2}-t^{3}$$
(iii) $$p(x)=x^{3}$$
(iv) $$p(x)=(x-1)(x+1)$$
Ans : (i) $$p(y)=y^{2}-y+1$$
$$p(0)=(0)^{2}-(0)+1=1$$
$$p(1)=(1)^{2}-(1)+1=1$$
$$p(2)=(2)^{2}-(2)+1=3$$

(ii) $$p(t)=2+t+2 t^{2}-t^{3}$$
$$p(0)=(0)^{2}-(0)+1=1$$
$$p(1)=(1)^{2}-(1)+1=1$$
$$p(2)=(2)^{2}-(2)+1=3$$

(iii) $$p(x)=x^{3}$$
$$p(0)=(0)^{3}=0$$
$$p(1)=(1)^{3}=1$$
$$p(2)=(2)^{3}=8$$

(iv) $$p(x)=(x-1)(x+1)$$
$$p(0)=(0-1)(0+1)=(-1)(1)=-1$$
$$p(1)=(1-1)(1+1)=0(2)=0$$
$$p(2)=(2-1)(2+1)=1(3)=3$$ 
Q.3: Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $$P(x)=3 x+1, x=-\frac{1}{3}$$
(ii) $$p(x)=5 x-\pi, x=\frac{4}{5}$$
(iii) $$p(x)=x^{2}-1, x=1,-1$$
(iv) $$p(x)=(x+1)(x-2), x=-1,2$$
(v) $$p(x)=x^{2}, x=0$$
(vi) $$p(x)=\left[x+m, x=-\frac{m}{l}\right.$$
(vii) $$P(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$$
(viii)  $$p(x)=2 x+1, x=\frac{1}{2}$$
Ans : (i)  If $$x=\frac{-1}{3}$$ is a zero of given polynomial $$p(x)=3 x+1$$ should be 0
Here $$p\left(\frac{-1}{3}\right)=3\left(\frac{-1}{3}\right)+1=-1+1=0$$
Therefore $$x=\frac{-1}{3}$$ is a zero of the given polynomial .

(ii) If $$x=\frac{4}{5}$$ is a zero of polynomial $$p(x)=5 x-\pi$$ then $$p\left(\frac{4}{5}\right)$$ should be 0.
Here $$p\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi=4-\pi$$
As $$p\left(\frac{4}{5}\right) \neq 0$$
Therefore $$x=\frac{4}{5}$$ is a zero of given polynomial .

(iii) If $$x=1$$ and $$x=-1$$ are zeros of polynomial $$p(x)=x^{2}-1$$ then $$p(1)$$  and $$p(-1)$$ should be 0.
Here $$p(1)=(1)^{2}-1=0$$ and $$p(-1)=(-1)^{2}-1=0$$
$$p(-1)=(-1)^{2}-1=0$$

(iv) If $$x=-1$$ and  $$x=2$$  is a zero of polynomial $$p(x)=(x+1)(x-2)$$ then $$p(-1) P(2)$$ should be 0.
Here $$p(-1)=(-1+1)(-1-2)=0(-3)=0$$ and  $$p(2)=(2+1)(2-2)=3(0)=0$$
Therefore $$x=-1 \text { and } x=2$$ are zeroes of the given polynomial

(v) If $$x=0 x=2$$ is a zero of polynomial $$p(x)=x^{2}$$ then $$p(0)$$ should be 0.
Here $$p(0)=(0)^{2}=0$$
Hence $$x=0$$ is a zero of the given polynomial .

(vi) If $$x=\frac{-m}{l}$$ is a zero of polynomial $$p(x)=|x+m$$ then $$p\left(\frac{-m}{l}\right)$$ should be 0.
Here, $$p\left(\frac{-m}{l}\right)=l\left(\frac{-m}{l}\right)+m=-m+m=0$$
Therefore $$x=-\frac{m}{l}$$ is a zero of the given polynomial .

(vii) If $$x=\frac{-1}{\sqrt{3}} \text { and } x=\frac{2}{\sqrt{3}}$$ are zeroes of polynomial $$p(x)=3 x^{2}-1$$ then $$p\left(\frac{-1}{\sqrt{3}}\right) \text { and } p\left(\frac{2}{\sqrt{3}}\right)$$ should be 0.
Here $$p\left(\frac{-1}{\sqrt{3}}\right)=3\left(\frac{-1}{\sqrt{3}}\right)^{2}-1=3\left(\frac{1}{3}\right)-1=1-1=0$$
And $$p\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{2}{\sqrt{3}}\right)^{2}-1=3\left(\frac{4}{3}\right)-1=4-1=3$$
Hence ,$$x=\frac{-1}{\sqrt{3}}$$ is a zero of the given polynomial .However, $$x=\frac{2}{\sqrt{3}}$$ is not a zero of the given polynomial .

(viii) If $$x=\frac{1}{2}$$ is a zero of polynomial $$p(x)=2 x+1$$ then $$p\left(\frac{1}{2}\right)$$ should be 0.
Here $$p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1=1+1=2$$
As  $$p\left(\frac{1}{2}\right) \neq 0$$
Therefore $$x=\frac{1}{2}$$ is not a zero of polynomial 
Q.4: Find the zero of the polynomial in each of the following cases

(i) $$p(x)=x+5$$
(ii) $$p(x)=x-5$$
(iii) $$p(x)=2 x+5$$
(iv) $$p(x)=3 x-2$$
(v) $$p(x)=3 x$$
(vi) $$p(x)=a x, a \neq 0$$
(vii) $$p(x)=c x+d, c \neq 0, c$$ are real numbers.
Ans : (i) Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
$$p(x)=x+5$$
$$p(x)=0$$
$$x+5=0$$
$$x=-5$$
Therefore for $$x=5$$ the value of the polynomial n is 0 and hence $$x=5$$ is a zero of the given polynomial .

(ii) $$p(x)=x-5$$
$$p(x)=0$$
x-5=0
x=5
Therefore for $$x=5$$ the value of the polynomial n is 0 and hence $$x=5$$ is a zero of the given polynomial .

(iii) $$p(x)=2 x+5$$
$$p(x)=0$$
$$2 x+5=0$$
$$2 x=-5$$
$$x=-\frac{5}{2}$$
Therefore for $$x=-\frac{5}{2}$$ the value of the polynomial n is 0 and hence $$x=\frac{-5}{2}$$ is a zero of the given polynomial .

(iv) $$p(x)=3 x-2$$
$$p(x)=0$$
$$3 x-2=0$$
$$x=\frac{2}{3}$$
Therefore for $$x=\frac{2}{3}$$ the value of the polynomial n is 0 and hence $$x=\frac{2}{3}$$ is a zero of the given polynomial .

(v) $$p(x)=3 x$$
$$p(x)=0$$
$$a x=0$$
$$x=0$$
Therefore for $$x=0$$ the value of the polynomial n is 0 and hence $$x=0$$ is a zero of the given polynomial .

(vi) $$p(x)=a x$$
$$p(x)=0$$
$$a x=0$$
$$x=0$$
Therefore for $$x=0$$ the value of the polynomial n is 0 and hence $$x=0$$ is a zero of the given polynomial .

(vii) $$p(x)=c x+d$$
$$p(x)=0$$
$$c x+d=0$$
$$x=\frac{-d}{c}$$
Therefore for $$x=\frac{-d}{c}$$ the value of the polynomial n is 0 and hence $$x=\frac{-d}{c}$$ is a zero of the given polynomial . 
Q.1: Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :

$$p(x)=x^{3}-3 x^{2}+5 x-3, \quad g(x)=x^{2}-2$$
$$p(x)=x^{4}-3 x^{2}+4 x+5, \quad g(x)=x^{2}+1-x$$
$$p(x)=x^{4}-5 x+6, \quad g(x)=2-x^{2}$$
Ans : (i) $$p(x)=x^{3}-3 x^{2}+5 x-3, g(x)=x^{2}-2$$ (ii)
$$p(x)=x^{4}-3 x^{2}+4 x+5=x^{4}+0 . x^{3}-3 x^{2}+4 x+5$$
$$q(x)=x^{2}+1-x=x^{2}-x+1$$ (iii)
$$p(x)=x^{4}-5 x+6=x^{4}+0 x^{2}-5 x+6$$
$$q(x)=2-x^{2}=-x^{2}+2$$ Q.2: Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
$$(i) t^{2}-3,2 t^{4}+3 t^{3}-2 t^{2}-9 t-12$$
(ii) $$x^{2}+3 x+1,3 x^{4}+5 x^{3}-7 x^{2}+2 x+2$$
(iii) $$x^{3}-3 x+1, x^{5}-4 x^{3}+x^{2}+3 x+1$$
Ans : (i)$$t^{2}-3,2 t^{4}+3 t^{3}-2 t^{2}-9 t-12$$
$$t^{2}-3=t^{2}+0 . t-3$$ Since the remainder is o,
Hence,$$t^{2}-3$$ is a factor of $$2 t^{4}+3 t^{3}-2 t^{2}-9 t-12$$
(ii)
$$\quad x^{2}+3 x+1,3 x^{4}+5 x^{3}-7 x^{2}+2 x+2$$ (iii)
$$x^{3}-3 x+1, x^{5}-4 x^{3}+x^{2}+3 x+1$$ Q.3: Obtain all other zeroes of $$3 x^{4}+6 x^{3}-2 x^{2}-10 x-5$$, if two of its zeroes are $$\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}$$.
Ans : $$p(x)=3 x^{4}+6 x^{3}-2 x^{2}-10 x-5$$
Since the two zeroes are $$\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}$$,
$$\therefore\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=\left(x^{2}-\frac{5}{3}\right)$$ is a factor of $$3 x^{4}+6 x^{3}-2 x^{2}-10 x-5$$.
Therefore, we divide the given polynimial by $$x^{2}-\frac{5}{3}$$. Hence, the zeroes of the given polynomial are $$\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1 \text { and }-1$$. 
Q.4: On dividing $$x^{3}-3 x^{2}+x+2$$ by a polynomial  g(x),  the quotient and remainder were x-2 and -2 x+4,  respectively. Find  g(x).
Ans : $$\begin{array}{l}{p(x)=x^{3}-3 x^{2}+x+2 \quad \text { (Dividend) }} \\ {g(x)=? \text { (Divisor) }} \\ {\text {Quotient }=(x-2)} \\ {\text {Remainder }=(-2 x+4)} \\ {\text {Dividend = Divisor * Quotient + Remainder }}\end{array}$$
$$\begin{array}{l}{x^{3}-3 x^{2}+x+2=g(x) \times(x-2)+(-2 x+4)} \\ {x^{3}-3 x^{2}+x+2+2 x-4=g(x)(x-2)} \\ {x^{3}-3 x^{2}+3 x-2=g(x)(x-2)} \\ {g(x) \text { is the quotient when we divide }\left(x^{3}-3 x^{2}+3 x-2\right) \text { by }(x-2)}\end{array}$$ Q.5: Give examples of polynomial p(x),g(x),q(x) and r(x) which satisfy the division algorithm and
(i) $$\operatorname{deg} p(x)=\operatorname{deg} q(x)$$
(ii) $$\operatorname{deg} q(x)=\operatorname{deg} r(x)$$
(iii) $$\operatorname{deg} r(x)=0$$
Ans : $$\begin{array}{l}{\text { According to the division algorithm, if } p(x) \text { and } g(x) \text { are two }} \\ {\text { polynomials with }} \\ {g(x) \neq 0, \text { then we can find polynomials } q(x) \text { and } r(x) \text { such that }} \\ {p(x)=g(x) \times q(x)+r(x)} \\ {\text { where } r(x)=0 \text { or degree of } r(x)<\text { degree of the variable in the }} \\ {\text { polynomial. }}\end{array}$$
$$\begin{array}{l}{\text { (i) deg } p(x)=\operatorname{deg} q(x)} \\ {\text { Degree of quotient will be equal to degree of dividend when divisor is }} \\ {\text { constant (i.e., when any polynomial is divided by a constant). }} \\ {\text { Let us assume the division of } 6 x^{2}+2 x+2 \text { by } 2 \text { . }}\end{array}$$

$$\begin{array}{l}{\text { Here, } p(x)=6 x^{2}+2 x+2} \\ {g(x)=2} \\ {q(x)=3 x^{2}+x+1 \text { and } r(x)=0} \\ {\text { Degree of } p(x) \text { and } q(x) \text { is the same i.e., } 2 \text { . }} \\ {\text { Checking for division algorithm, }} \\ {p(x)=g(x) \times q(x)+r(x)}\end{array}$$
$$\begin{array}{l}{6 x^{2}+2 x+2=2\left(3 x^{2}+x+1\right)} \\ {=6 x^{2}+2 x+2} \\ {\text { Thus, the division algorithm is satisfied. }}\end{array}$$

$$\begin{array}{l}{\text { (ii) deg } q(x)=\operatorname{deg} r(x)} \\ {\text { Let us assume the division of } x^{3}+x \text { by } x^{2},} \\ {\text { Here, } p(x)=x^{3}+x} \\ {g(x)=x^{2}} \\ {q(x)=x \text { and } r(x)=x} \\ {\text { Clearly, the degree of } q(x) \text { and } r(x) \text { is the same i.e., } 1 .}\end{array}$$
$$\begin{array}{l}{\text { Checking for division algorithm, }} \\ {p(x)=g(x) \times q(x)+r(x)} \\ {x^{3}+x=\left(x^{2}\right) \times x+x} \\ {x^{3}+x=x^{3}+x} \\ {\text { Thus, the division algorithm is satisfied. }}\end{array}$$

$$\begin{array}{l}{\text { (iii) deg } r(x)=0} \\ {\text { Degree of remainder will be } 0 \text { when remainder comes to a constant. }} \\ {\text { Let us assume the division of } x^{3}+1 \text { by } x^{2} \text { . }} \\ {\text { Here, } p(x)=x^{3}+1} \\ {g(x)=x^{2}}\end{array}$$

$$\begin{array}{l}{q(x)=x \text { and } r(x)=1} \\ {\text { Clearly, the degree of } r(x) \text { is } 0 \text { . }} \\ {\text { Checking for division algorithm, }} \\ {p(x)=g(x) \times q(x)+r(x)} \\ {x^{3}+1=\left(x^{2}\right) \times x+1} \\ {x^{3}+1=x^{3}+1} \\ {\text { Thus, the division algorithm is satisfied. }}\end{array}$$ 
Q.1: Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) $$2 x^{3}+x^{2}-5 x+2 ; \frac{1}{2}, 1,-2$$
(ii) $$x^{3}-4 x^{2}+5 x-2 ; 2,1,1$$
Ans : $$\begin{array}{l}{\text { (i) } P(x)=2 x^{3}+x^{2}-5 x+2} \\ {\text { Zeroes for this polynomial are } \frac{1}{2}, 1,-2} \\ {P\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{2}-5\left(\frac{1}{2}\right)+2} \\ {=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2} \\ {=0}\end{array}$$

$$\begin{array}{l}{p(1)=2 \times 1^{3}+1^{2}-5 \times 1+2} \\ {\quad=0} \\ {p(-2)=2(-2)^{3}+(-2)^{2}-5(-2)+2} \\ {\quad=-16+4+10+2=0} \\ {\text { Therefore, } \frac{1}{2}, 1, \text { and }-2 \text { are the zeroes of the given polynomial. }} \\ {\text { Comparing the given polynomial with } a x^{3}+b x^{2}+c x+d, \text { we obtain } a=2,} \\ {b=1, c=-5, d=2}\end{array}$$

$$\begin{array}{l}{\text { We can take } \alpha=\frac{1}{2}, \beta=1, y=-2} \\ {\alpha+\beta+\gamma=\frac{1}{2}+1+(-2)=-\frac{1}{2}=\frac{-b}{a}} \\ {\alpha \beta+\beta \gamma+\alpha \gamma=\frac{1}{2} \times 1+1(-2)+\frac{1}{2}(-2)=\frac{-5}{2}=\frac{c}{a}} \\ {\alpha \beta \gamma=\frac{1}{2} \times 1 \times(-2)=\frac{-1}{1}=\frac{-(2)}{2}=\frac{-d}{a}} \\ {\text { Therefore, the relationship between the zeroes and the coefficients is }} \\ {\text { verified. }}\end{array}$$

$$\begin{array}{l}{\text { (ii) } p(x)=x^{3}-4 x^{2}+5 x-2} \\ {\text { Zeroes for this polynomial are } 2,1,1 \text { . }} \\ {p(2)=2^{3}-4\left(2^{2}\right)+5(2)-2} \\ {=8-16+10-2=0} \\ {p(1)=1^{3}-4(1)^{2}+5(1)-2} \\ {=1-4+5-2=0}\end{array}$$

$$\begin{array}{l}{\text { Therefore, } 2,1,1 \text { are the zeroes of the given polynomial. }} \\ {\text { Comparing the given polynomial with } a x^{3}+b x^{2}+c x+d, \text { we obtain } a=1} \\ {b=-4, c=5, d=-2}\end{array}$$
$$\begin{array}{l}{\text {Verification of the relationship between zeroes and coefficient of the }} \\ {\text { given polynomial }} \\ {\text { Sum of zeroes } 2+1+1=4=\frac{-(-4)}{1}=\frac{-b}{a}} \\ {\text { Multiplication of zeroes taking two at a time }=(2)(1)+(1)(1)+(2)(1)}\end{array}$$
$$\begin{array}{l}{=2+1+2=5^{=\frac{(5)}{1}=\frac{c}{a}}} \\ {\text { Multiplication of zeroes }=2 \times 1 \times 1=2^{=\frac{-(-2)}{1}=\frac{-d}{a}}} \\ {\text { Hence, the relationship between the zeroes and the coefficients is }} \\ {\text { verified. }}\end{array}$$ 
Q.2: Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time. and the product of its zeroes as 2, -7, -14 respectively.
Ans : $$\begin{array}{l}{\text { Let the polynomial be } a x^{3}+b x^{2}+c x+d \text { and the zeroes be } \alpha, \beta, \text { and } \gamma} \\ {\text { It is given that }} \\ {\alpha+\beta+\gamma=\frac{2}{1}=\frac{-b}{a}} \\ {\alpha \beta+\beta \gamma+\alpha \gamma=\frac{-7}{a}=\frac{c}{a}} \\ {\alpha \beta \gamma=\frac{-14}{1}=\frac{-d}{a}} \\ {\text { If } a=1, \text { then } b=-2, c=-7, d=14} \\ {\text { Hence, the polynomial is } x^{3}-2 x^{2}-7 x+14.}\end{array}$$
Q.3: If the zeroes of the polynomial $$x^{3}-3 x^{2}+x+1 \text { are } a-b, a, a+b, \text { find } a \text { and } b.$$
Ans : $$p(x)=x^{3}-3 x^{2}+x+1$$
Zeroes are a-b,a+a+b
comparing the given polynomial with $$p x^{3}+q x^{2}+r x+t$$,we obtain
p=1,q=-3,r=1,t=1
sum of zeroes = a-b+a+a+b
$$\frac{-q}{p}=3 a$$
$$\frac{-(-3)}{1}=3 a$$
3=3a
a=1
The zeroes are 1-b,1,1+b
Multiplication of zeroes =1(1-b)(1+b)
$$\frac{-t}{p}=1-b^{2}$$
$$\frac{-1}{1}=1-b^{2}$$
$$1-b^{2}=-1$$
$$1+1=b^{2}$$
$$b=\pm \sqrt{2}$$ 
Q.4: If two zeroes of the polynomial $$x^{4}-6 x^{3}-26 x^{2}+138 x-35 \text { are } 2 \pm \sqrt{3},$$find other zeroes.
Ans : Given that $$2+\sqrt{3}$$ and $$2-\sqrt{3}$$ are zeroes of the given polynomial.
Therefore, $$(x-2-\sqrt{3})(x-2+\sqrt{3})=x^{2}+4-4 x-3$$
=$$x^{2}-4 x+1$$ is a factor of the given polynomial
For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing $$x^{4}-6 x^{3}-26 x^{2}+138 x-35 \text { by } x^{2}-4 x+1$$ Q.5: $$\begin{array}{l}{\text { If the polynomial } x^{4}-6 x^{3}+16 x^{2}-25 x+10 \text { is divided by another polynomial } x^{2}-2 x+k} \\ {\text { the remainder comes out to be } x+a, \text { find } k \text { and } a .}\end{array}$$
Ans : $$\begin{array}{l}{\text { By division algorithm, }} \\ {\text { Dividend = Divisor } \times \text { Quotient + Remainder }} \\ {\text { Dividend - Remainder = Divisor } \times \text { Quotient }} \\ {x^{4}-6 x^{3}+16 x^{2}-25 x+10-x-a=x^{4}-6 x^{3}+16 x^{2}-26 x+10-a_{\text { will be perfectly }}} \\ {\text { divisible by } x^{2}-2 x+k \text { . }} \\ {\text { Let us divide } x^{4}-6 x^{3}+16 x^{2}-26 x+10-a b y x^{2}-2 x+k}\end{array}$$ Q.1: Find the remainder when x3 + 3x2 + 3x + 1 is divided by

(i) $$x+1$$
(ii) $$x=\frac{1}{2}$$
(iii) $$x$$
(iv) $$x+\pi$$
(v) $$5+2 x$$
Ans : (i) (ii) (iii) (iv) (v) Q.2: Find the remainder when $$x^{3}+3 x^{2}+3 x+1$$  is divided by x – a.
Ans : Q.3: Check whether 7 + 3x is a factor of $$3 x^{3}+7 x$$
Ans : Q.1: Determine which of the following polynomials has (x + 1) a factor :

(i) $$x^{3}+x^{2}+x+1$$
(ii) $$x^{4}+x^{2}+x^{2}+x+1$$
(iii) $$x^{4}+3 x^{3}+3 x^{2}+x+1$$
(iv) $$x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$$
Ans : (i) If $$(x+1)$$ is a factor of $$p(x)=x^{3}+x^{2}+x+1$$ then $$p(-1)$$ must be zero , otherwise $$(x+1)$$ is not of $$p(x)$$ .
$$p(x)=x^{3}+x^{2}+x+1$$
$$p(-1)=(-1)^{3}+(-1)^{2}+(-1)+1$$
$$=-1+1-1-1=0$$
Hence $$x+1$$ is not a factor of this polynomial .

(ii) If $$(x+1)$$ is a factor of $$p(x)=x^{4}+x^{3}+x^{2}+x+1$$ then $$p(-1)$$ must be zero , otherwise $$(x+1)$$ is not of $$p(x)$$ .
$$p(x)=x^{4}+x^{3}+x^{2}+x+1$$
$$p(-1)=(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)+1$$
$$=1-1+1-1+1=1$$
Therefore $$x+1$$ is not a factor of this polynomial .

(iii) If $$(x+1)$$ is a factor of polynomial $$p(x)=x^{4}+3 x^{3}+3 x^{2}+x+1$$ then $$p(-1)$$ must be 0 otherwise $$(x+1)$$ is not a factor of this polynomial .
$$p(-1)=(-1)^{4}+3(-1)^{3}+3(-1)^{2}+(-1)+1$$
$$=1-3+3-1+1=1$$
As $$p(-1) \neq 0$$
Therefore, x+1 is not a factor of this polynomial .

(iv)  If $$x+1$$ is a factor of polynomial $$p(x)=x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$$ then $$p(-1)$$ must be 0 otherwise $$(x+1)$$ is not a factor of this polynomial .
$$p(-1)=(-1)^{3}-(-1)^{2}-(2+\sqrt{2})(-1)+\sqrt{2}$$
$$=-1-1+2+\sqrt{2}+\sqrt{2}$$
$$=2 \sqrt{2}$$
As $$p(-1) \neq 0$$
Therefore, x+1 is not a factor of this polynomial . 
Q.2: Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) $$p(x)=2 x^{3}+x^{2}-2 x-1, g(x)=x+1$$
(ii)  $$p(x)=x^{3}+3 x^{2}+3 x+1, g(x)=x+2$$
(iii) $$P(x)=x^{3}-4 x^{2}+x+6, g(x)=x-3$$
Ans : If $$g(x)=x+1$$  is a factor of the given polynomial p(x) then p(-1) must be zero .
$$p(x)=2 x^{3}+x^{2}-2 x-1$$
$$p(-1)=2(-1)^{3}+(-1)^{2}-2(-1)-1$$
$$=2(-1)+1+2-1=0$$
Hence $$g(x)=x+1$$ is a factor of the polynomial .

(ii) If $$g(x)=x+2$$ is a factor of the given polynomial . $$p(x)$$ then $$p(-2)$$ must be 0.
$$p(x)=x^{3}+3 x^{2}+3 x+1$$
$$p(-2)=(-2)^{3}+3(-2)^{2}+3(-2)+1$$
$$=-8+12-6+1$$
$$=-1$$
As $$p(-2) \neq 0$$
Hence $$g(x)=x+2$$ is not a factor of the given polynomial.

(iii) If $$g(x)=x-3$$ is a factor of the given polynomial . $$p(x)$$ then $$p(3)$$ must be 0.
$$p(x)=x^{3}-4 x^{2}+x+6$$
$$p(3)=(3)^{3}-4(3)^{2}+3+6$$
$$=27-36+9=0$$
Hence $$g(x)=x-3$$ is not a factor of the given polynomial. 
Q.3: Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i) $$p(x)=x^{2}+x+k$$
(ii) $$p(x)=2 x^{2}+k x+\sqrt{2}$$
(iii) $$p(x)=k x^{2}-\sqrt{2} x+1$$
(iv) $$p(x)=k x^{2}-3 x+k$$
Ans : (i) $$p(x)=x^{2}+x+k$$
$$p(1)=0$$
$$\Rightarrow(1)^{2}+1+k=0$$
$$\Rightarrow 2+k=0$$
$$\Rightarrow k=-2$$
Therefore the value of k is -2

(ii) $$p(x)=2 x^{2}+k x+\sqrt{2}$$
$$p(1)=0$$
$$\Rightarrow 2(1)^{2}+k(1)+\sqrt{2}=0$$
$$\Rightarrow 2+k+\sqrt{2}=0$$
$$\Rightarrow k=-2-\sqrt{2}=-(2+\sqrt{2})$$
Therefore ,the value of k is $$-(2+\sqrt{2})$$

(iii) $$p(x)=k x^{2}-\sqrt{2} x+1$$
$$p(1)=0$$
$$\Rightarrow k(1)^{2}-\sqrt{2}(1)+1=0$$
$$\Rightarrow k-\sqrt{2}+1=0$$
$$\Rightarrow k=\sqrt{2}-1$$
Therefore the value of k is  $$\sqrt{2}-1$$

(iv) $$p(x)=k x^{2}-3 x+k$$
$$\Rightarrow p(1)=0$$
$$\Rightarrow k(1)^{2}-3(1)+k=0$$
$$\Rightarrow k-3+k=0$$
$$\Rightarrow 2 k-3=0$$
$$\Rightarrow k=\frac{3}{2}$$
Therefore the value of k is $$\frac{3}{2}$$ 
Q.4: Factorise :

(i) $$12 x^{2}-7 x+1$$
(ii) $$2 x^{2}+7 x+3$$
(iii) $$p(x)=k x^{2}-\sqrt{2} x +1$$
(iv) $$3 x^{2}-x=4$$
Ans : (i) $$12 x^{2}-7 x+1$$
We can find two numbers such that $$p q=12 \times 1=12$$ and $$p+q=-7$$ .They are $$p=$$  4 and $$q=-3$$
Here $$12 x^{2}-7 x+1=12 x^{2}-4 x-3 x+1$$
$$=4 x(3 x-1)-1(3 x-1)$$
$$=(3 x-1)(4 x-1)$$

(ii) $$2 x^{2}+7 x+3$$
We can	find two numbers such that $$p q=2 \times 3=6$$ and $$p+q=7$$
They are $$p=6$$ and $$q=1$$ .
Here $$2 x^{2}+7 x+3=2 x^{2}+6 x+x+3$$
$$=2 x(x+3)+1(x+3)$$
$$=(x+3)(2 x+1)$$

(iii) $$6 x^{2}+5 x-6$$
We can	find two numbers such that $$p q=-36$$ and $$p+q=5$$
They are $$p=9$$  and $$q=-4$$
Here,
$$6 x^{2}+5 x-6=6 x^{2}+9 x-4 x-6$$
$$=3 x(2 x+3)-2(2 x+3)$$
$$=(2 x+3)(3 x-2)$$

(iv) $$3 x^{2}-x-4$$
We can find two numbers such that $$p q=3 \times(-4)=-12$$
And $$p+q=-1$$
They are $$p=-4 \text { and } q=3$$
Here,
$$3 x^{2}-x-4=3 x^{2}-4 x+3 x-4$$
$$=x(3 x-4)+1(3 x-4)$$
$$=(3 x-4)(x+1)$$ 
Q.5: Factorise :

(i) $$x^{3}-2 x^{2}-x+2$$
(ii)  $$x^{2}-3 x^{2}-9 x-5$$
(iii) $$x^{3}+13 x^{2}+32 x+20$$
(iv) $$2 y^{3}+y^{2}-2 y-1$$
Ans : (i) Let $$p(x)=x^{3}-2 x^{2}-x+2$$
All the factors of 2 have to be considered .These are $$\pm 1, \pm 2$$
By trial method
$$p(2)=(2)^{3}-2(2)^{2}-2+2$$
$$=8-8-2+2=0$$
Therefore $$(x-2)$$ is a factor of polynomial $$p(x)$$
Let us find out the quotient on dividing $$x^{3}-2 x^{2}-x+2$$  by p(x)
By long division , It is known that
Dividend = Divisor * Quotient + Remainder
$$\therefore x^{3}-2 x^{2}-x+2=(x+1)\left(x^{2}-3 x+2\right)+0$$
$$=(x+1)\left[x^{2}-2 x-x+2\right]$$
$$=(x+1)[x(x-2)-1(x-2)]$$
$$=(x+1)(x-1)(x-2)$$
$$=(x-2)(x-1)(x+1)$$

(ii) Let $$p(x)=x^{3}-3 x^{2}-9 x-5$$
All the factors of 5 have to considered .These are $$\pm 1, \pm 5$$
By Trial method
$$p(-1)=(-1)^{3}-3(-1)^{2}-9(-1)-5$$
$$=-1-3+9-5=0$$
Therefore, $$x+1$$ is a factor of this polynomial .
Let us find the quotient on dividing $$x^{3}+3 x^{2}-9 x-5 \text { by } x+1$$
By long division, It is known that
Dividend = Divisor * Quotient + Remainder
$$\therefore x^{3}-3 x^{2}-9 x-5=(x+1)\left(x^{2}-4 x-5\right)+0$$
$$=(x+1)\left(x^{2}-5 x+x-5\right)$$
$$=(x+1)[(x(x-5)+1(x-5)]$$
$$=(x+1)(x-5)(x+1)$$
$$=(x-5)(x+1)(x+1)$$

(iii) Let $$p(x)=x^{3}+13 x^{2}+32 x+20$$
All the factors of 20 have to be considered Some of them are $$\pm 1$$
$$\pm 2, \pm 4, \pm 5 \ldots \ldots$$
By Trial Method
$$p(-1)=(-1)^{3}+13(-1)^{2}+32(-1)+20$$
$$=-1+13-32+20$$
$$=33-33=0$$
As $$p(-1)$$ is zero , therefore $$x+1$$ is a factor of his polynomial $$p(x)$$
Let us find the quotient on dividing $$x^{3}+13 x^{2}+32 x+20 \text { by }(x+1)$$
By long division , It is known that
Dividend = Divisor * Quotient + Remainder
$$x^{3}+13 x^{2}+32 x+20=(x+1)\left(x^{2}+12 x+20\right)+0$$
$$=(x+1)\left(x^{2}+10 x+2 x+20\right)$$
$$=(x+1)[x(x+10)+2(x+10)]$$
$$=(x+1)(x+10)(x+2)$$
$$=(x+1)(x+2)(x+10)$$

(iv) Let $$p(y)=2 y^{3}+y^{2}-2 y-1$$
By Trial Method
$$p(1)=2(1)^{3}+(1)^{2}-2(1)-1$$
$$=2+1-2-1=0$$
Therefore $$y-1$$ is a factor of this polynomial .
Let us find the quotient on dividing $$2 y^{3}+y^{2}-2 y-1 \text { by } y-1$$ Q.1: Use suitable identities to find the following products

(i) $$(x+4)(x+10)$$
(ii) $$(x+8)(x-10)$$
(iii) $$(3 x+4)(3 x-5)$$
(iv) $$\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)$$
(v) $$(3-2 x)(3+2 x)$$
Ans : (i) By using the identity $$(x+a)(x+b)=x^{2}+(a+b) x+a b$$
$$(x+4)(x+10)=x^{2}+(4+10) x+4 \times 10$$
$$=x^{2}+14 x+40$$

(ii) By using the identity $$(x+a)(x+b)=x^{2}+(a+b) x+a b$$
$$(x+8)(x-10)=x^{2}+(8-10) x+(8)(-10)$$
$$=x^{2}-2 x-80$$

(iii) By using the identity $$(3 x+4)(3 x-5)=9\left(x+\frac{4}{3}\right)\left(x-\frac{5}{3}\right)$$
$$(x+a)(x+b)=x^{2}+(a+b) x+a b$$
$$9\left(x+\frac{4}{3}\right)\left(x-\frac{5}{3}\right)=9\left[x^{2}+\left(\frac{4}{3}-\frac{5}{3}\right) x+\left(\frac{4}{3}\right)\left(-\frac{5}{3}\right)\right]$$
$$9\left[x^{2}-\frac{1}{3} x-\frac{20}{9}\right]$$
$$9 x^{2}-3 x-20$$

(iv) By using the identity $$(x+y)(x-y)=x^{2}-y^{2}$$
$$\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}$$
$$y^{4}-\frac{9}{4}$$

(v) By using the identity $$(x+y)(x-y)=x^{2}-y^{2}$$
$$(3-2 x)(3+2 x)=(3)^{2}-(2 x)^{2}$$
$$9-4 x^{2}$$ 
Q.2: Evaluate the following products without multiplying directly:

(i) $$103 \times 107$$
(ii) $$95 \times 96$$
(iii) $$104 \times 96$$
Ans : (i) $$103 \times 107=(100+3)(100+7)$$
$$=(100)^{2}+(3+7) 100+(3)(7)$$
By using the identity $$(x+a)(x+b)=x^{2}+(a+b) x+a b$$ where
$$x=100, a=3, \text { and } b=7 ]$$
$$=10000+1000+21$$
$$=11021$$

(ii) $$95 \times 96=(100-5)(100-4)$$
$$=(100)^{2}+(-5-4) 100+(-5)(-4)$$
By using the identity $$(x+a)(x+b)=x^{2}+(a+b) x+a b$$ where
$$x=100, a=-5, \text { and } b=-4 ]$$
$$=10000-900+20$$
$$=9120$$

(iii) $$104 \times 96=(100+4)(100-4)$$
$$(100)^{2}-(4)^{2}\left[(x+y)(x-y)=x^{2}-y^{2}\right]$$
$$10000-16$$
$$9984$$ 
Q.3: Factorise the following using appropriate identities:

(i) $$9 x^{2}+6 x y+y^{2}$$
(ii) $$4 y^{2}-4 y+1$$
(iii)  $$x^{2}-\frac{y^{2}}{100}$$
Ans : (i) $$9 x^{2}+6 x y+y^{2}=(3 x)^{2}+2(3 x)(y)+(y)^{2}$$
$$=(3 x+y)(3 x+y)$$

(ii)  $$4 y^{2}-4 y+1=(2 y)^{2}-2(2 y)(1)+(1)^{2}$$
$$=(2 y-1)(2 y-1) \quad\left[x^{2}-2 x y+y^{2}=(x-y)^{2}\right]$$

(iii)  $$x^{2}-\frac{y^{2}}{100}=x^{2}-\left(\frac{y}{10}\right)^{2}$$
$$=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right) \quad\left[x^{2}-y^{2}=(x+y)(x-y)\right]$$ 
Q.4: Expand each of the following, using suitable identities:

(i) $$(x+2 y+4 z)^{2}$$
(ii) $$(2 x-y+z)^{2}$$
(iii) $$(-2 x+3 y+2 z)^{2}$$
(iv) $$(3 a-7 b-c)^{2}$$
(v) $$(-2 x+5 y-3 z)^{2}$$
(vi) $$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$$
Ans : (i) It is known that ,
$$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$$
$$(x+2 y+4 z)^{2}=x^{2}+(2 y)^{2}+(4 z)^{2}+2(x)(2 y)+2(2 y)(4 z)+2(4 z)(x)$$
$$(x+2 y+4 z)^{2}=x^{2}+(2 y)^{2}+(4 z)^{2}+2(x)(2 y)+2(2 y)(4 z)+2(4 z)(x)$$
$$=x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 x z$$

(ii) $$(2 x-y+z)^{2}=(2 x)^{2}+(-y)^{2}+(z)^{2}+2(2 x)(-y)+2(-y)(z)+2(z)(2 x)$$
$$=4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z$$

(iii) $$(-2 x+3 y+2 z)^{2}$$
$$=(-2 x)^{2}+(3 y)^{2}+(2 z)^{2}+2(-2 x)(3 y)+2(3 y)(2 z)+2(2 z)(-2 x)$$
$$=4 x^{2}+9 y^{2}+4 z^{2}-12 x y+12 y z-8 x z$$

(iv) $$(3 a-7 b-c)^{2}$$
$$=(3 a)^{2}+(-7 b)^{2}+(-c)^{2}+2(3 a)(-7 b)+2(-7 b)(-c)+2(-c)(3 a)$$
$$=9 a^{2}+49 b^{2}+c^{2}-42 a b+14 b c-6 a c$$

(v) $$(-2 x+5 y-3 z)^{2}$$
$$=(-2 x)^{2}+(5 y)^{2}+(-3 z)^{2}+2(-2 x)(5 y)+2(5 y)(-3 z)+2(-3 z)(-2 x)$$
$$=4 x^{2}+25 y^{2}+9 z^{2}-20 x y-30 y z+12 x z$$

(vi) $$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$$
$$\left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2\left(\frac{1}{4} a\right)(1)$$
$$=\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a$$ 
Q.5: Factorise:

(i) $$4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$$
(ii) $$2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$$
Ans : It is known that
$$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$$
(i) $$4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$$
$$=(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+2(2 x)(3 y)+2(3 y)(-4 z)+2(2 x)(-4 z)$$
$$=(2 x+3 y-4 z)^{2}$$
$$=(2 x+3 y-4 z)(2 x+3 y-4 z)$$

(ii) $$2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$$
$$(-\sqrt{2} x)^{2}+(y)^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z)$$
$$(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}$$
$$(-\sqrt{2} x+y+2 \sqrt{2} z)(-\sqrt{2} x+y+2 \sqrt{2} z)$$ 
Q.6: Write the following cubes in expanded form
(i) $$(2 x+1)^{3}$$
(ii) $$(2 a-3 b)^{3}$$
(iii) $$\left[\frac{3}{2} x+1\right]^{3}$$
(iv) $$\left[x-\frac{2}{3} y\right]^{3}$$
Ans : It is known that
$$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$$
And $$(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$$
(i) $$(2 x+1)^{3}=(2 x)^{3}+(1)^{3}+3(2 x)(1)(2 x+1)$$
$$= 8 x^{3}+1+6 x(2 x+1)$$
$$= 8 x^{3}+1+12 x^{2}+6 x$$
$$=8 x^{3}+12 x^{2}+6 x+1$$

$$\begin{array}{l}{(ii )(2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)(2 a-3 b)} \\ {=8 a^{3}-27 b^{3}-18 a b(2 a-3 b)} \\ {=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}}\end{array}$$

(iii) $$\begin{array}{l}{\left[\frac{3}{2} x+1\right]^{3}=\left[\frac{3}{2} x\right]^{3}+(1)^{3}+3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x+1\right)} \\ {=\frac{27}{8} x^{3}+1+\frac{9}{2} x\left(\frac{3}{2} x+1\right)} \\ {=\frac{27}{8} x^{3}+1+\frac{27}{4} x^{2}+\frac{9}{2} x} \\ {=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1}\end{array}$$
(iv) $$\begin{array}{l}{\left[x-\frac{2}{3} y\right]^{3}=x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)} \\ {=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)} \\ {=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}}\end{array}$$ 
Q.7: Evaluate the following using suitable identities:
$$(\mathrm{i})(99)^{3}(\mathrm{ii})(102)^{3} \text { (iii) }(998)^{3}$$
Ans : $$\begin{array}{l}{\text { It is known that, }} \\ {(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)} \\ {\text { and }(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)}\end{array}$$

$$\begin{array}{l}{(\mathrm{i})(99)^{3}=(100-1)^{3}} \\ {=(100)^{3}-(1)^{3}-3(100)(1)(100-1)} \\ {=1000000-1-300(99)} \\ {=1000000-1-29700} \\ {=970299}\end{array}$$
$$\begin{array}{l}{(i i)(102)^{3}=(100+2)^{3}} \\ {=(100)^{3}+(2)^{3}+3(100)(2)(100+2)} \\ {=1000000+8+600(102)} \\ {=1000000+8+61200} \\ {=1061208}\end{array}$$

$$\begin{array}{l}{\left(\text { iii) }(998)^{3}=(1000-2)^{3}\right.} \\ {=(1000)^{3}-(2)^{3}-3(1000)(2)(1000-2)} \\ {=1000000000-8-6000(998)} \\ {=1000000000-8-5988000} \\ {=1000000000-5988008} \\ {=994011992}\end{array}$$ 
Q.8: Factorise each of the following:
$$\begin{array}{l}{\text { (i) } 8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}\left (\text {(ii) } 8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\right.} \\ {\text { (iii) } 27-125 a^{3}-135 a+225 a^{2} \text { (iv) } 64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}} \\ {(v)}{\quad 27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p}\end{array}$$
Ans : $$\begin{array}{l}{\text { It is known that, }} \\ {(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}} \\ {\text { and }(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}} \\ {\text { (i) } 8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}} \\ {=(2 a)^{3}+(b)^{3}+3(2 a)^{2} b+3(2 a)(b)^{2}} \\ {=(2 a+b)^{3}} \\ {=(2 a+b)(2 a+b)(2 a+b)}\end{array}$$

$$\begin{array}{l}{\left(\text { ii) } 8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\right.} \\ {=(2 a)^{3}-(b)^{3}-3(2 a)^{2} b+3(2 a)(b)^{2}} \\ {=(2 a-b)^{3}} \\ {=(2 a-b)(2 a-b)(2 a-b)}\end{array}$$

$$\begin{array}{l}{\text { (iii) } 27-125 a^{3}-135 a+225 a^{2}} \\ {=(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2}} \\ {=(3-5 a)^{3}} \\ {=(3-5 a)(3-5 a)(3-5 a)}\end{array}$$

$$\begin{array}{l}{\left(\text { iv) } 64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\right.} \\ {=(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}} \\ {=(4 a-3 b)^{3}} \\ {=(4 a-3 b)(4 a-3 b)(4 a-3 b)}\end{array}$$

$$\begin{array}{l}{\quad(v)^{27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p}} \\ {=(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)^{2}\left(\frac{1}{6}\right)+3(3 p)\left(\frac{1}{6}\right)^{2}} \\ {=\left(3 p-\frac{1}{6}\right)^{3}} \\ {=\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)}\end{array}$$ 
Q.9: Verify:
\begin{array}{l} { (i) } {x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)} \\ {( ii )}{x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)}\end{array}
Ans : \begin{aligned} \text { (i) It is known that, } \\(x+y)^{3} &=x^{3}+y^{3}+3 x y(x+y) \\ x^{3}+y^{3} &=(x+y)^{3}+3 x y(x+y) \\ &=(x+y)\left[(x+y)^{2}-3 x y\right] \\ &=(x+y)\left(x^{2}+y^{2}+2 x y-3 x y\right) \\ &=(x+y)\left(x^{2}+y^{2}+2 x y-3 x y\right) \\ &=(x+y)\left(x^{2}-x y+y^{2}\right) \end{aligned}
\begin{aligned} \text { (ii) It is known that, } \\(x-y)^{3} &=x^{3}-y^{3}-3 x y(x-y) \\ x^{3}-y^{3} &=(x-y)^{3}+3 x y(x-y) \\ &=(x-y)\left[(x-y)^{2}+3 x y\right] \\ &=(x-y)\left(x^{2}+y^{2}+3 x y\right) \\ &=(x-y)\left(x^{2}+y^{2}+x y\right) \\ &=(x-y)\left(x^{2}+y^{2}+x y\right) \\ &=(x-y)\left(x^{2}+x y+y^{2}\right) \end{aligned} 
Q.10: $$\begin{array}{l}{\text { Factorise each of the following: }} \\ {\text { (i) } 27 y^{3}+125 z^{3}} \\ {\text { (ii) } 64 m^{3}-343 n^{3}} \\ {\text { [Hint: See question] } 9 .}\end{array}$$
Ans : $$\begin{array}{l}{\text { (i) }^{27 y^{3}+125 z^{3}}} \\ {=(3 y)^{3}+(5 z)^{3}} \\ {=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right] \quad\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]} \\ {=(3 y+5 z)\left[9 y^{2}+25 z^{2}-15 y z\right]}\end{array}$$
$$\begin{array}{l}{\text { (ii) } 64 m^{3}-343 n^{3}} \\ {=(4 m)^{3}-(7 n)^{3}} \\ {=(4 m-7 n)\left[(4 m)^{2}+(7 n)^{2}+(4 m)(7 n)\right] \quad\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]} \\ {=(4 m-7 n)\left[16 m^{2}+49 n^{2}+28 m n\right]}\end{array}$$ 
Q.11: Factorise: $$27 x^{3}+y^{3}+z^{3}-9 x y z$$
Ans : $$\begin{array}{l}{\text { It is known that, }} \\ {x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)} \\ {\therefore 27 x^{3}+y^{3}+z^{3}-9 x y z} \\ {=(3 x)^{3}+(y)^{3}+(z)^{3}-3(3 x)(y)(z)(z)-z(3 x)} \\ {=(3 x+y+z)\left[9 x^{2}+y^{2}+z^{2}-3 x y-y z-3 x z\right]}\end{array}$$
Q.12: Verify that $$x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$$
Ans : $$\begin{array}{l}{\text { It is known that, }} \\ {x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)} \\ {=\frac{1}{2}(x+y+z)\left[2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x\right]} \\ {=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+2 z^{2}+z^{2}-2 y z\right)+\left(x^{2}+z^{2}-2 z x\right)} \\ {=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]}\end{array}$$
Q.13: If $$x+y+z=0, \text { show that } x^{3}+y^{3}+z^{3}-3 x y z$$
Ans : $$\begin{array}{l}{\text { It is known that, }} \\ {x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)} \\ {\text { Put } x+y+z=0} \\ {x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)} \\ {x^{3}+y^{3}+z^{3}-3 x y z=0} \\ {x^{3}+y^{3}+z^{3}=3 x y z}\end{array}$$
Q.14: $$\begin{array}{l}{\text { Without actually calculating the cubes, find the value of each of the following: }} \\ {(1)(-12)^{3}+(7)^{3}+(5)^{3}} \\ {\text { (ii) }(28)^{3}+(-15)^{3}+(-13)^{3}}\end{array}$$
Ans : ( i )
$$\begin{array}{l}{(-12)^{3}+(7)^{3}+(5)^{3}} \\ {\text { Let } x=-12, y=7, \text { and } z=5} \\ {\text { It can be observed that, }} \\ {x+y+z=-12+7+5=0} \\ {\text { It is known that if } x+y+z=0 \text { , then }} \\ {x^{3}+y^{3}+z^{3}=3 x y z} \\ {\therefore(-12)^{3}+z^{3}=3 x y z} \\ {\therefore(-12)^{3}+z^{3}=3 x y z} \\ {=-1260}\end{array}$$
(ii) $$\begin{array}{l}{(28)^{3}+(-15)^{3}+(-13)^{3}} \\ {\text { Let } x=28, y=-15, \text { and } z=-13} \\ {\text { It can be observed that, }} \\ {x+y+z=28+(-15)+(-13)=28-28=0} \\ {\text { It is known that if } x+y+z=0, \text { then }} \\ {x^{3}+y^{3}+z^{3}=3 x y z}\end{array}$$
\begin{aligned} \therefore(28)^{3}+(-15)^{3}+(-13)^{3} &=3(28)(-15)(-13) \\ &=16380 \end{aligned} 
Q.15: Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
$$\frac{\left[\text { Area: } 25 \mathrm{a}^{2}-35 a+12\right.}{I} \quad \frac{\text { Area: } 35 y^{2}+13 y-12}{II}$$
Ans : $$\begin{array}{l}{\text { Area }=\text { Length } \times \text { Breadth }} \\ {\text { The expression given for the area of the rectangle has to be factorised. One of its }} \\ {\text { factors will be its length and the other will be its breadth. }} \\ {\text { (i) } 25 a^{2}-35 a+12=25 a^{2}-15 a-20 a+12} \\ {=5 a(5 a-3)-4(5 a-3)} \\ {=(5 a-3)(5 a-4)} \\ {\text { Therefore, possible length }=5 a-3} \\ {\text { And, possible breadth = } 5 a-4}\end{array}$$
$$\begin{array}{l}{\text { (ii) } 35 y^{2}+13 y-12=35 y^{2}+28 y-15 y-12} \\ {=7 y(5 y+4)-3(5 y+4)} \\ {=(5 y+4)(7 y-3)} \\ {\text { Therefore, possible length }=5 y+4} \\ {\text { And, possible breadth }=7 y-3}\end{array}$$ 
Q.16: What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
$$\frac{\text { Volume: } 3 x^{2}-12 x}{I} \quad \frac{\text { Volume: } 12 k y^{2}+8 k y-20 k}{II}$$
Ans : Volume of cuboid = Length $$\times$$ Breadth $$\times$$ Height The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.

$$\begin{array}{l}{3 x^{2}-12 x=3 x(x-4)} \\ {\text { One of the possible solutions is as follows. }} \\ {\text { Length }=3, \text { Breadth }=x, \text { Height }=x-4}\end{array}$$

$$\begin{array}{l}{\text { (ii) }^{12 k y^{2}+8 k y-20 k=4 k\left(3 y^{2}+2 y-5\right)}} \\ {=4 k\left[3 y^{2}+5 y-3 y-5\right]} \\ {=4 k[y(3 y+5)-1(3 y+5)]} \\ {=4 k(3 y+5)(y-1)} \\ {\text { One of the possible solutions is as follows. }} \\ {\text { Length }=4 k_{k} \text { , Breadth }=3 y+5, \text { Height }=y-1}\end{array}$$ 

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