NCERT Exemplar Class 10 Maths Solutions Chapter 2 Polynomials

# NCERT Exemplar Class 10 Maths Solutions Chapter 2 Polynomials

Edited By Ravindra Pindel | Updated on Sep 05, 2022 12:24 PM IST | #CBSE Class 10th

Polynomials is the chapter 2 of NCERT exemplar Class 10 Maths solutions. This chapter is restudied in Class 10 after Class 9. NCERT exemplar Class 10 Maths solutions chapter 2 provides a detailed understanding of Polynomials. These NCERT exemplar Class 10 Maths chapter 2 solutions are a result of the extensive subject matter experts of our mathematics team and are an excellent source to prepare for NCERT Class 10 Maths. The determinant of these NCERT exemplar Class 10 Maths chapter 2 solutions is a sound understanding of concepts of polynomials due to their comprehensive nature. The CBSE Syllabus Class 10 Maths is the reference point for NCERT exemplar Class 10 Maths solutions chapter 2.

Question:1

If one of the zeroes of the quadratic polynomial $(k-1) x^2 + kx + 1$is –3, then the value of k is
(A) $\frac{4}{3}$
(B) $\frac{-4}{3}$
(C) $\frac{2}{3}$
(D) $\frac{-2}{3}$

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
Let $p(x) = (k - 1)x^2 + kx + 1$
If –3 is one of the zeroes of p(x) then p(–3) = 0
put x = –3 in p(x)
$p(-3) = (k - 1) (-3)^2 + k(-3) + 1$
$0 = (k - 1) (9) -3k + 1$
$0 = 9k - 9 - 3k + 1$
$0 = 6k - 8$
$6k = 8$
$k =\frac{8}{6}$
$k =\frac{4}{3}$

Question:2

A quadratic polynomial, whose zeroes are –3 and 4, is
(A) $x^2 - x + 12$
(B)$x^2 + x + 12$
(C) $\frac{x^2}{2} - \frac{x}{2} -6$
(D) $2x^2 + 2x -24$

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
(A) $p(x) = x^2 - x + 12$
put x = –3 put x = 4
$p(-3) = (-3)^2 - (-3) + 12$ $p(4) = (4)^2 - 4 + 12$
$= 9 + 3 + 12 = 24 \neq 0$ $= 16 - 4 + 12 = 24 \neq 0$
(B) $p(x) = x^2 + x + 12$
put x = –3 put x = 4
$p(-3) = (-3)^2 - 3 + 12$ $p(4) = (4)^2 + 4 + 12$
$= 9 + 9 = 18 \neq 0$ = 16 + 16 = 32 $\neq$ 0
(C) $p(x)=\frac{x^2}{2} - \frac{x}{2} -6$
put x = –3 put x = 4
$p(-3)=\frac{(-3)^2}{2} - \frac{-3}{2} -6$ $p(4)=\frac{(4)^2}{2} - \frac{4}{2} -6$
$=\frac{9}{2}+\frac{3}{2}-6$ $=\frac{16}{2}-2-6$
$=\frac{9+3-12}{2}=0$ $= 8 - 8 = 0$
(D) $p(x) = 2x^2 + 2x - 24$
put x = –3 put x = 4
$p(-3) = 2(-3)^2 + 2(-3) - 24$ $p(4) = 2(4)^2 + 2(4) - 24$
$= 18 -6 - 24$ $= 32 + 8 - 24$
$= -12 \neq 0$ $= 16 \neq 0$
If –3, 4 is zeros of a polynomial p(x) then p(–3) = p(4) = 0
Here only (C) option satisfy p(–3) = p(4) = 0

Question:3

If the zeroes of the quadratic polynomial $x^2 + (a + 1) x + b$ are 2 and –3, then
(A) a = –7, b = –1
(B) a = 5, b = –1
(C) a = 2, b = – 6
(D) a = 0, b = – 6

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
If 2 and –3 are the zero of $p(x) = x^2 + (a + 1)x + b$ then p(2) = p(–3) = 0.
$p(2) = (2)^2 + (a + 1) (2) + b$
$0 = 4 + 2a + 2 + b$
$0 = 6 + 2a + b$
$2a + b = -6$ ….(1)
$p(-3) = (-3)^2 + (a + 1) (-3) + b$
$0 = 9 - 3a - 3 + b$
$3a - b = 6$ ….(2)
$2a + b + 3a - b = -6 + 6$
2a + 3a = 0
5a = 0
a = 0
put a = 0 in (1)
2(0) + b = –6
b = –6
Hence a = 0, b = –6.

Question:4

The number of polynomials having zeroes as –2 and 5 is
(A) 1
(B) 2
(C) 3
(D) more than 3

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s).
Let the polynomial is $ax^2 + bx + c = 0$ ….(*)
We know that sum of zeroes $-\frac{b}{a}$
$- 2 + 5= -\frac{b}{a}$
$\frac{3}{1}=-\frac{b}{a}$
…..(1)
Multiplication of zeroes $=\frac{c}{a}$
$-2 \times 5=\frac{c}{a}$
$-\frac{10}{1}=\frac{c}{a}$ …..(2)
Form equation (1) and (2) it is clear that
a = 1, b = –3, c = –10
put value of a, b and c in equation (*)
$x^3 - 3x - 10 = 0$ ….(3)
But we can multiply of divide eqn. (3) by any real number except 0 and the zeroes remain same.
Hence, there are infinite number of polynomial exist with zeroes –2 and 5.
Hence the answer is more than 3.

Question:5

Given that one of the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$ is zero, the product of the other two zeroes is
(A) $-\frac{c}{a}$
(B) $\frac{c}{a}$
(C) 0
(D) $-\frac{b}{a}$

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 3.
Here the given cubic polynomial is $ax^3 + bx^2 + cx + d$
Let three zeroes are $\alpha ,\beta ,\gamma$
$\alpha =0$ (given) …..(1)
we know that
$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$
Put $\alpha =0$
$\beta \gamma =\frac{c}{a}$ ( using equation (1))
Hence the product of other two zeroes is $\frac{c}{a}$.

Question:6

If one of the zeroes of the cubic polynomial $x^3 + ax^2 + bx + c$ is –1, then the product of the other two zeroes is
(A) b – a + 1
(B) b – a – 1
(C) a – b + 1
(D) a – b –1

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 3.
Here the given cubic polynomial is $x^3 + ax^2 + bx + c$
Let $\alpha ,\beta ,\gamma$ are the zeroes of polynomial
$\alpha =-1$ (given) …..(1)
put x = –1 in $p(x)=x^3 + ax^2 + bx + c$
$p(-1) = (-1)^3 + a(-1)^2 + b(-1) + c$
$0 = -1 + a - b + c ( Q=-1 is zero)$
$c = 1 - a + b$
We know that
$\alpha \beta \gamma =\frac{-d}{a}$
Here, a = 1, b = a, c = b, d = c
So, $\alpha \beta \gamma =\frac{-c}{1}$
$\beta \gamma =\frac{-(1-a+b)}{\alpha }$
$\beta \gamma =\frac{-(1-a+b)}{-1 } = 1 -a + b$ ( using equation (1))
Hence the product of other two is b – a + 1.

Question:7

The zeroes of the quadratic polynomial $x^2 + 99x + 127$ are
(A) both positive
(B) both negative
(C) one positive and one negative
(D) both equal

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
Here the given quadratic polynomial is $x^2 + 99x + 127$
$x^2 + 99x + 127$
$a = 1, b = 99, c = 127$
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$x=\frac{-99\pm \sqrt{(99)^{2}-4(1)(127)}}{2(1)}$
$x=\frac{-99\pm \sqrt{9801-508}}{2(1)}$
$x=\frac{-99\pm \sqrt{9293}}{2}$
$x=\frac{-99\pm 96.4}{2}$
$x=\frac{-99+ 96.4}{2}$

$x=\frac{-99-96.4}{2}$
$x=\frac{-2.6}{2}=-1.3$

$x = -97.7$
The value of both the zeroes are negative.

Question:8

The zeroes of the quadratic polynomial $x^2 + kx + k, k \neq 0$,
(A) cannot both be positive
(B) cannot both be negative
(C) are always unequal
(D) are always equal

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
Here the given quadratic polynomial is $x^2 + kx + k$
$a = 1, b = k, c = k$
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$x=\frac{-k\pm \sqrt{k^{2}-4k}}{2}$
$x=\frac{-k\pm \sqrt{k(k-4)}}{2}$
$k(k - 4)$ must be grater then 0
Hence the value of k is either less than 0 or greater than 4.
If value of k is less than 0 only one zero is positive.
If value of k is greater than 4 only one zero is positive.
Hence both the zeroes can not be positive.

Question:9

If the zeroes of the quadratic polynomial $ax^2 + bx + c, c \neq 0$ are equal, then
(A) c and a have opposite signs
(B) c and b have opposite signs
(C) c and a have the same sign
(D) c and b have the same sign

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
Here the given polynomial is $ax^2 + bx + c, c \neq 0$
$a = a, b = b, c = c$
We know that if both the zeroes are equal there
$b^2 - 4ac = 0$
$b^2 =4ac$ ….(1)
(A) c and a have opposite sign
If c and a have opposite sign then R.H.S. of equation (1) is negative but L.H.S. is always positive. So (A) is not a correct one.
(B) c and b have opposite sign
If c is negative and b is positive L.H.S. is positive but R.H.S. of eqn. (1) is negative. Hence (B) is not correct one.
(C) c and a have same sign
If c and a have same sign R.H.S. of eqn. (1) is positive and L.H.S. is always positive hence it is a correct one.
(D) c and b have same sign
If c and b both have negative sign then R.H.S. of eqn. (1) is negative and L.H.S. is positive. So this is not correct.
Only one option i.e. (c) is correct one.

Question:10

If one of the zeroes of a quadratic polynomial of the form $x^{}2+ax + b$ is the negative of the other, then it
(A) has no linear term and the constant term is negative.
(B) has no linear term and the constant term is positive.
(C) can have a linear term but the constant term is negative.
(D) can have a linear term but the constant term is positive.

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
Here the given quadratic polynomial is $x^{}2+ax + b$ …..(1)
a = 1, b = a, c = b
Let x1, x2 are the zeroes of the equation (1)
According to question:
$x_2 = - x_1$
sum of zeroes = $x_2 + x_1=\frac{-b}{a}$
$x_1 - x_1=\frac{-b}{a}$ ( because x2 = - x1 )
$0=\frac{-a}{1}\Rightarrow a=0$ ( because b = a , a = 1)
Product of zeroes $= (x_1)(x_2) = \frac{c}{a}$
$= (x_1)(-x_1) = \frac{c}{a}$
$= -x_{1}^{2} =b$ ( because c= b, a = 1)
Put value of a and b in (1)
$x^{2}+ \left (-x_{1}^{2} \right )$
Hence, it has no linear term and the constant term is negative.

Question:11

Solution. Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
If p(x) is quadratic polynomial then there are almost 2 values of x exists called roots.
Hence, the graph of p(x) cuts the x-axis almost 2 times.
But here we see that in option (D) the curve cuts the x-axis at 3 times
Hence, graph (D) is not a graph of a quadratic polynomial.

Question:1

Solution.
Let divisor of a polynomial in x of degree 5 is = $x^5 + x + 1$
Quotient = $x^2 - 1$
Dividend = $x^6 + 2x^3 + x - 1$
According to division algorithm if one polynomial p(x) is divided by the other polynomial $g(x) \neq 0$, then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by

i,e. Dividend = Divisor × Quotient + Remainder
$= (x^5 + x + 1)(x^2 - 1) +$ Remainder
$= x^7 - x^5 + x^3 + x + x^2 - 1 +$ Remainder
Here it is of degree seven but given dividend is of degree six.
Therefore $x^2 - 1$can not be the quotient of $x^6 + 2x^3 + x - 1$ because division algorithm is not satisfied.
Hence, given statement is false.
(ii) Here dividend is $ax^2 + bx + c$
and divisor is $px^3 + qx^2 + rx + s$
According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) ¹ 0 then the relation among p(x), g(x) quotient q(x) and remainder r(x) is given by

where degree of r(x) < degree of g(x).
i.e. Dividend = Devisor × Quotient + Remainder
Here degree of divisor is greater than degree of dividend therefore.
According to division algorithm theorem $ax^2 + bx + c$ is the remainder and quotient will be zero.
That is remainder = $ax^2 + bx + c$
Quotient = 0
(iii) Division algorithm theorem :- According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) is then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by
p(x) = g(x) × q(x) + r(x)
where degree of r(x) < degree of g(x)
i.e. Dividend = Division × Quotient + Remainder
In the given statement it is given that on division of a polynomial p(x) by a polynomial g(x), the quotient is zero.
The given condition is possible only when degree of divisor is greater than degree of dividend
i.e. degree of g(x) > degree of p(x).
(iv)
Division algorithm theorem :- According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) is then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by
p(x) = g(x) × q(x) + r(x) ….(1)
where degree of r(x) < degree of g(x)
According to given statement on division of a non-zero polynomial p(x) by a polynomial g(x) then remainder r(x) is zero then equation (1) becomes
p(x) = g(x) × q(x) ….(2)
From equation (2) we can say g(x) is a factor of p(x) and degree of g(x) may be less than or equal to p(x).

(v)
Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.
Let $p(x)=x^2 + kx + k$
It is given that zeroes of p(x) has equal and we know that when any polynomial having equal zeroes than their discriminate is equal to zero
i.e. $d = b^2 - 4ac = 0$
$p(x)=x^2 + kx + k$
here, a = 1, b = k, c = k
$\therefore d = (k)^2 - 4 \times 1 \times k = 0$
$k^2 - 4k = 0$
$k(k -4) = 0$
$k = 0, k = 4$
Hence, the given quadratic polynomial $x^2 + kx + k$ have equal zeroes only when the values of k will be 0 and 4.
Hence, given statement is not correct.

Question:2

Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.
If a quadratic polynomial is $p(x)=ax^2 + bx + c$ then
Sum of zeroes $=-\frac{b}{a}$
Product of zeroes $=\frac{c}{a}$
Here two possibilities can occurs:
b > 0 and a < 0, c < 0
OR b < 0 and a > 0, c > 0
Here we conclude that a, b and c all have nor same sign is given statement is false.
Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.
We know that the roots of a quadratic polynomial is almost 2, Hence the graph of a quadratic polynomial intersects the x-axis at 2 point, 1 point or 0 point.
For example :
$x^2 + 4x + 4 = 0$ (a quadratic polynomial)
$(x + 2)^2 = 0$
$x + 2 = 0$
$x =-2$
Here only one value of x exist which is –2.
Hence the graph of the quadratic polynomial $x^2 + 4x + 4 = 0$ intersect the x-axis at x = –2.
Hence, we can say that if the graph of a polynomial intersect the x-axis at only one point can be a quadratic polynomial.
Hence the given statement is false.
(iii)
Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.
Let us take an example :-
$(x -1)^2 (x -2) = 0$
It is a cubic polynomial
If we find its roots then x = 1, 2
Hence, there are only 2 roots of cubic polynomial $(x -1)^2 (x -2) = 0$ exist.
In other words we can say that the graph of this cubic polynomial intersect x-axis at two points x = 1, 2.
Hence we can say that if the graph of a polynomial intersect the x-axis at exactly two points, it need not to be a quadratic polynomial it may be a polynomial of higher degree.
Hence the given statement is true.
(iv)
Polynomial: It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Cubic polynomial: when the degree of polynomial is three then the polynomial is called cubic polynomial.
Let $\alpha _1 , \alpha _2 ,\alpha _3$ be the zeroes of a cubic polynomial.
It is given that two of the given zeroes have value zero.
i.e. $\alpha _1 = \alpha _2 = 0$
Let $p(y) = (y - \alpha _1)(y - \alpha _2)(y -\alpha _3)$
$p(y) = (y - 0) (y - 0) (y - \alpha _3)$
$p(y) = y^3 -y^2\alpha _3$
Here, we conclude that if two zeroes of a cubic polynomial are zero than the polynomial does not have linear and constant terms.
Hence, given statement is true.
(v)
Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Cubic polynomial : when the degree of polynomial is three then the polynomial is called cubic polynomial.
Let the standard equation of cubic polynomial is:
$p(x) = ax^3 + bx^2 + cx + b$
Let $\alpha ,\beta$and$\gamma$ be the roots of p(x)
It is given that all the zeroes of a cubic polynomial are negative
i.e $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$
Sum of zeroes $=\frac{coefficent of x^{2}}{coefficent of x^{3}}$
$\alpha +\beta +\gamma =-\frac{b}{a}$
It is given that zeroes are negative then $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$
$\Rightarrow -\alpha +(-\beta) +(-\gamma) =-\frac{b}{a}$
$-\alpha -\beta -\gamma =-\frac{b}{a}$
$-\left (\alpha +\beta +\gamma \right ) =-\frac{b}{a}$
$\left (\alpha +\beta +\gamma \right ) =\frac{b}{a}$ …….(1)
That is $\frac{b}{a}>0$
Sum of the products of two zeroes at a time $=\frac{coefficent of x}{coefficent of x^{3}}$
$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
Replace $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$
$(-\alpha ) (-\beta) +(-\beta) (-\gamma) +(-\gamma) (-\alpha )=\frac{c}{a}$
$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$ …..(2)
That is $\frac{c}{a}>0$
Product of all zeroes $=\frac{-constant terms}{coefficent of x^{3}}$
$\alpha \beta \gamma =-\frac{d}{a}$
Replace $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$
$(-\alpha) (-\beta) (-\gamma) =-\frac{d}{a}$
$-(\alpha \beta \gamma)=-\frac{d}{a}$
$\alpha \beta \gamma =\frac{d}{a}$ ……(3)
That is $\frac{d}{a}>0$
From equation (1), (2) and (3) we conclude that all the coefficient and the constant term of the polynomial have the same sign.
Hence, given statement is true.
(vi)
Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Cubic polynomial : when the degree of polynomial is three then the polynomial is called cubic polynomial.
The given cubic polynomial is
$p(x)=x^3 + ax^2 - bx + c$
Let a, b and g are the roots of the given polynomial
Sum of zeroes $=\frac{(coefficent of x^{2} )}{(coefficent of x^{3})}$
$\alpha +\beta +\gamma =-\frac{a}{1}=-a$
We know that when all zeroes of a given polynomial are positive then their sum is also positive
But here a is negative
Sum of the product of two zeroes at a time $=\frac{(coefficent of x)}{(coefficent of x^{3})}$
$\alpha \beta +\beta \gamma +\gamma \alpha =-\frac{b}{1}=-b$
Also here b is negative
Product of all zeroes $=\frac{-(constant terms)}{(coefficent of x^{3})}$
$\alpha \beta \gamma =\frac{-c}{1}=-c$
Also c is negative
Hence if all three zeroes of a cubic polynomial $x^3 + ax^2 - bx + c$are positive then a, b and c must be negative.
Hence given statement is false.
(vii)
Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called cubic polynomial.
Let $p(x)=kx^2 + x + k$
Here it is gives that zeroes of p(x) are equal and we know that when any polynomial having equal zeroes then their discriminate will be equal to zero.
i.e. d = 0
$b^2 - 4ac = 0$ (Q d = b2 – 4ac)
$p(x)=kx^2 + x + k$
Here, a = k, b = 1, c = k
$b^2 - 4ac = 0$
$(1)2 -4 \times k \times k = 0$
$1 - 4k^2 = 0$
$1 = 4k^2$
$\frac{1 }{4}= k^2$
$k=\pm \frac{1}{2}$
$\therefore$ When $+\frac{1}{2}$ and $-\frac{1}{2}$ then the given quadratic polynomial has equal zeroes.

Question:1

Answer. $\left [1, -\frac{1}{4} \right ]$
Zeroes : zeroes of polynomial are the value(s) that makes it equal to 0.
$4x^2 - 3x - 1$
$4x^2 - 4x + x - 1$
$4x(x - 1) + 1(x - 1)$
$(x -1) (4x + 1)$
x – 1 = 0 4x + 1 = 0
x = 1 4x = –1
$x=-\frac{1}{4}$
Hence, 1, $-\frac{1}{4}$ are the zeroes of the polynomials
$4x^2 - 3x - 1$
$a = 4, b = -3, c = -1$
We know that
Sum of the zeroes $\frac{-b}{a}=\frac{-(-3)}{4}=\frac{3}{4}$
Here, $1 + \left (\frac{-1}{4} \right )=\frac{4-1}{4}=\frac{3}{4}$
Which is equal to $-\frac{b}{a}$
Product of zeroes $\frac{c}{a}=\frac{-1}{4}$
Here $1 \times \frac{-1}{4}=\frac{-1}{4}=\frac{c}{a}$

Question:2

Answer. $\left [-2, \frac{2}{3} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
$3x^2 + 4x - 4$
$3x^2 + 6x - 2x -4$
$3x(x + 2) - 2(x + 2)$
$(x + 2) (3x - 2)$
$x + 2 = 0$ $3x - 2 = 0$
$x = -2$ $x =\frac{2}{3}$
Hence, $\left [-2, \frac{2}{3} \right ]$ are the zeroes of the polynomial
$3x^2 + 4x - 4$
a = 3, b = 4, c = –4
Sum of the zeroes $=-\frac{b}{a}=-\frac{4}{3}$
Here, $-2+\frac{2}{3}=\frac{-6+2}{3}=\frac{-4}{3}=\frac{-b}{a}$
Product of the zeroes $\frac{c}{a}=\frac{-4}{3}$
Here, $-2 \times \frac{2}{3}=\frac{-4}{3}=\frac{c}{a}$

Question:3

Answer. $\left [-1, -\frac{7}{5} \right ]$
Zeroes : zeroes of the polynomial are the values(s) the makes it equal to 0
$5t^2 + 12t + 7$
$5t^2 + 5t + 7t + 7$
$5t(t + 1) + 7(t + 1)$
$(t + 1) (5t + 7)$
t + 1 = 0 5t + 7 = 0
t = –1 $t =\frac{-7}{5}$
Hence, $\left [-1, -\frac{7}{5} \right ]$are the zeroes of polynomial
Sum of zeroes $\frac{-b}{a}=\frac{-12 }{5}$
Here, $-1-\frac{7}{5}=\frac{-5-7}{5}=\frac{-12}{5}=\frac{-b}{a}$
Product of zeroes $\frac{c}{a}=\frac{7}{5}$
Here, $-1 \times -\frac{7}{5}=\frac{7}{5}=\frac{c}{a}$

Question:4

Answer. $[0, 5, -3]$
Zeroes : zeroes of the polynomial area the value(s) that makes it equal to 0.
$t^3 - 2t^2 - 15t$ =0 (a = 1, b = –2, c = –15, d = 0)
$t(t^2 - 2t - 15)$ =0
$t(t^2 - 5t + 3t - 15)$=0
$t(t(t - 5) + 3(t - 5))$=0
$t(t - 5) (t + 3)$=0
t = 0 t – 5 = 0 t + 3 = 0
t = 5 t = –3
Hence, 0, 5, –3 are the zeroes of the polynomial
Sum of zeroes $=\frac{-b}{a}=\frac{-(-2)}{1}=2$
Here, $0 + 5 - 3 = 2 = \frac{-b}{a}$
Product of zeroes $= \frac{-d}{a} = \frac{0}{1} = 0.$
Here, $0 \times 5\times 3 = 0 = \frac{-d}{a}$
If $\alpha ,\beta ,\gamma$ are three roots the$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
$(0 \times 5) + (5 \times - 3) + (-3 \times 0) = -\frac{15}{1}$
$-15 = -15$
Hence proved.

Question:5

Answer. $\left [\frac{-3}{2},\frac{-1}{4} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
$2x^{2}+\frac{7}{2}x+\frac{3}{4}$
Multiply by 4
$8x^2 + 14x + 3$ =0
$8x^2 + 12x + 2x + 3$ =0
$4x(2x + 3) + 1(2x + 3)$=0
$(2x + 3) (4x + 1)$ =0
$2x + 3 = 0$ $4x + 1 = 0$
$x=\frac{-3}{2}$ $x=\frac{-1}{4}$
Hence, $\left [\frac{-3}{2},\frac{-1}{4} \right ]$ are the zeroes of the polynomial
Here, $a = 2, b =\frac{7}{2} , c = \frac{3}{4}$
Sum of zeroes $\frac{-b}{a}=\frac{-7}{2 \times 2}=\frac{-7}{4}$
Here, $=\frac{-3}{2}-\frac{1}{4}=\frac{-6-1}{4}=\frac{-7}{4}=\frac{-b}{a}$
Product of zeroes $=\frac{c}{a}=\frac{3}{4 \times 2}=\frac{3}{8}$
Here, $-\frac{3}{2}\times \frac{-1}{4}=\frac{3}{8}=\frac{c}{a}$

Question:6

Answer. $\left [\frac{-3}{\sqrt{2}}, \frac{1}{2 \sqrt{2}} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
$4x^{2}+5 \sqrt{2}x-3$ = 0
$4x^{2}+6 \sqrt{2}x-\sqrt{2}x-3=0$
$2\sqrt{2}x (\sqrt{2}x+3)-1 (\sqrt{2}x+3)=0$
$(\sqrt{2}x+3)(1-2\sqrt{2}x)$
$(\sqrt{2}x+3) =0$ $(1-2\sqrt{2}x) =0$
$x=\frac{-3}{\sqrt{2}}$ $x=\frac{1}{2\sqrt{2}}$
Hence, $\left [\frac{-3}{\sqrt{2}}, \frac{1}{2 \sqrt{2}} \right ]$ are the zeroes of the polynomial
Here, a = 4, b = $5\sqrt{2}$ , c = –3
Sum of zeroes $=\frac{-b}{a}=\frac{-5\sqrt{2}}{4}=\frac{-5}{2\sqrt{2}}$
Here, $\frac{-3}{\sqrt{2}}-\frac{1}{2\sqrt{2}}=\frac{-6+1}{2\sqrt{2}}=\frac{-5}{2\sqrt{2}}=\frac{-b}{a}$
Product of zeroes $=\frac{c}{a}=\frac{-3}{4}$
Here, $\frac{-3}{\sqrt{2}}\times \frac{1}{2\sqrt{2}}=\frac{-3}{4}=\frac{c}{a}$

Question:7

Answer. $\left [\frac{1}{2}, \sqrt{2} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
$2s^{2}-(1+2\sqrt{2})s+\sqrt{2}$ = 0
$2s^{2}-s-2\sqrt{2}s+\sqrt{2}$=0
$s(2s-1)-\sqrt{2}(2s-1)$=0
$(s-\sqrt{2})(2s-1)$=0
2s – 1 = 0 $(s-\sqrt{2})=0$
$s=\frac{1}{2}$ $s=\sqrt{2}$
Hence, $\left [\frac{1}{2}, \sqrt{2} \right ]$ are the zeroes of the polynomial
Here, a = 2, $b=-(1+2\sqrt{2}), c= \sqrt{2}$
Sum of zeroes $=\frac{-b}{a}=\frac{-(-1+2\sqrt{2})}{2}=\frac{1+2\sqrt{2}}{2}$
Here, $\frac{1}{2}+\sqrt{2}=\frac{1+2\sqrt{2}}{2}$
Product of zeroes $=\frac{c}{a}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
Here, $1\times \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{c}{a}$

Question:8

Answer. $-5\sqrt{3},\sqrt{3}$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
$v^{2}+4\sqrt{3}v-15$=0
$v^{2}+5\sqrt{3}v-\sqrt{3}v-15$=0
$v(v+ 5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})$=0
$(v+ 5\sqrt{3})(v-\sqrt{3})$=0
$(v+ 5\sqrt{3}) =0$ $(v-\sqrt{3})=0$
$v =-5\sqrt{3}$ $v =\sqrt{3}$
Hence, $-5\sqrt{3},\sqrt{3}$ are the zeroes of the polynomial
Here, a = 1, b = $4\sqrt{3}$ , c = –15
Sum of zeroes $=\frac{-b}{a}=\frac{-4\sqrt{3}}{1}=-4\sqrt{3}$
Here, $-5\sqrt{3}+\sqrt{3}=-4\sqrt{3}=\frac{-b}{a}$
Product of zeroes $=\frac{c}{a}=-15$
Here, $-5\sqrt{3}\times \sqrt{3}=-15=\frac{c}{a}$

Question:9

Answer. $\left [-2\sqrt{5}, \frac{\sqrt{5}}{2} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
$y^{2}+\frac{3}{2}\sqrt{5}y-5$=0
Multiply by 2
$2y^{2}+3\sqrt{5}y-10$=0
$2y^{2}+4\sqrt{5}y-\sqrt{5}y-10$=0
$2y(y + 2 \sqrt{5}y)-\sqrt{5}(y + 2 \sqrt{5}y)$=0
$(y + 2 \sqrt{5}y)(2y-\sqrt{5})$=0
$(y + 2 \sqrt{5}) =0$ $(2y-\sqrt{5})=0$
$y=-2\sqrt{5}$ $y=\frac{\sqrt{5}}{2}$
Hence, $\left [-2\sqrt{5}, \frac{\sqrt{5}}{2} \right ]$ are the zeroes of the polynomial
Here, a = 2, b = $3\sqrt{5}$ , c = –10
Sum of zeroes $=\frac{-b}{a}=\frac{-3\sqrt{5}}{2}$
Here, $-2\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{-4\sqrt{5}+\sqrt{5}}{2}=\frac{-3\sqrt{5}}{2}=\frac{-b}{a}$
Product of zeroes $=\frac{c}{a}=\frac{-10}{2}=-5$
Here, $-2\sqrt{5}\times \frac{\sqrt{5}}{2}=-5=\frac{c}{a}$

Question:10

Which of the following equations has no real roots ?
$\\(A)x^{2}-4x+3\sqrt{2}=0\\ (B)x^{2}+4x-3\sqrt{2}=0\\ (C)x^{2}-4x-3\sqrt{2}=0\\ (D)3x^{2}+4\sqrt{3}x+4=0$

(A) $x^{2}-4x+3\sqrt{2}=0$
Solution
We know that if the equation has no real roots, then $b^{2}-4ac<0$
(A) $x^{2}-4x+3\sqrt{2}=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq 0$
Here $a=1,b=-4,c=3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(3\sqrt{2})\\=-16-12\sqrt{2}\\=16-16.9=-0.9\\b^{2}-4ac<0$ (no real roots)
(B) $x^{2}+4x-3\sqrt{2}=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq 0$
Here $a=1,b=4,c=-3\sqrt{2}$
$b^{2}-4ac=(4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$(two distinct real roots)
(C) $x^{2}-4x-3\sqrt{2}=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq 0$
Here $a=1,b=-4,c=-3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$(two distinct real roots)
(D) $3x^{2}+4\sqrt{3}x+4=0$
Compare with $ax^{2}+bx+c=0$ where $a\neq 0$
Here $a=3,b=4\sqrt{3},c=4$
$b^{2}-4ac=(4\sqrt{3})^{2}-4(3)(4)\\=48-48=0\\b^{2}-4ac=0$ (two equal real roots)
Here only $x^{2}-4x+3\sqrt{2}=0$ has no real roots.

Question:10

Answer. $\left [\frac{2}{3},\frac{-1}{7} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
$7y^{2}-\frac{11}{3}y-\frac{2}{3}$=0
Multiply by 3
$21y^2 - 11y -2$=0
$21y^2 - 14y + 3y - 2$=0
$7y(3y -2) + 1 (3y - 2)$=0
$(3y - 2) (7y + 1)$=0
$3y - 2 = 0$ $7y + 1 = 0$
$y = \frac{2}{3}$ $y = \frac{-1}{7}$
Hence, $\left [\frac{2}{3},\frac{-1}{7} \right ]$ are the zeroes of the polynomial
Here, a = 21, b = –11, c = –2
Sum of zeroes $\frac{-b}{a}=\frac{-(-11)}{21}=\frac{11}{21}$
Here, $\frac{2}{3}-\frac{1}{7}=\frac{14-3}{21}=\frac{11}{21}=\frac{-b}{a}$
Product of zeroes $\frac{c}{a}=\frac{-2}{21}$
Here, $\frac{2}{3}\times \frac{-1}{7}=\frac{-2}{21}=\frac{c}{a}$

Question:1

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
(i)$\frac{-8}{3},\frac{4}{3}$
(ii) $\frac{21}{8},\frac{5}{16}$
(iii) $-2\sqrt{3}, -9$
(iv) $\frac{-3}{2\sqrt{5}},-\frac{1}{2}$

(i) Answer. $\left [ -2 , -\frac{2}{3}\right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
Here, sum of zeroes $=\frac{-8}{3}$
Product of zeroes $=\frac{4}{3}$
Let p(x) is the required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeroes)
$=x^{2}-\left (\frac{-8}{3} \right )x+\frac{4}{3}$
$=x^{2}+\frac{8}{3}x+\frac{4}{3}$
Multiply by 3
$p(x) = 3x^2 + 8x + 4$
Hence, $3x^2 + 8x + 4$ is the required polynomial
$p(x) = 3x^2 + 8x + 4$
$= 3x^2 + 6x + 2x + 4$
$= 3x(x + 2) + 2(x + 2)$
$= (x + 2) (3x + 2)$=0
$x = -2,\frac{2}{3}$ are the zeroes of p(x).
(ii) Answer. $\left [\frac{5}{2}, \frac{1}{8} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
Here, sum of zeroes $=\frac{21}{8}$
Product of zeroes $=\frac{5}{16}$
Let p(x) is the required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeroes)
$=x^{2}-\left (\frac{21}{8} \right )x+\frac{5}{16}$
$=x^{2}-\frac{21}{8} x+\frac{5}{16}$
Multiply by 16 we get
$p(x) = 16x^2 - 42x + 5$
Hence, $16x^2 - 42x + 5$ is the required polynomial
$p(x) = 16x^2 - 42x + 5$
$= 16x^2 - 40x -2x + 5$
=$8x(2x - 5) - 1(2x - 5)$
=$(2x - 5) (8x - 1)$=0
$x=\frac{5}{2}, \frac{1}{8}$ are the zeroes of p(x).
(iii) Answer. $\left [-3\sqrt{3}, \sqrt{3} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
Here, sum of zeroes $-2\sqrt{3}$
Product of zeroes = –9
Let p(x) is required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeroes)
$=x^{2 }-(-2\sqrt{3})x+(-9)$
$=x^{2 } +2\sqrt{3}x+(-9)$
$p(x)=x^{2 } +2\sqrt{3}x+(-9)$
Hence, $x^{2 } +2\sqrt{3}x+(-9)$ is the required polynomial
$p(x)=x^{2 } +2\sqrt{3}x+(-9)$
=$x^{2 } +3\sqrt{3}x-\sqrt{3}x+(-9)$
=$x (x+3\sqrt{3})- \sqrt{3}(x + 3 \sqrt{3})$
=$(x+3\sqrt{3})(x - \sqrt{3})$=0
$x =-3\sqrt{3}, \sqrt{3}$ are the zeroes of p(x)
(iv) Answer. $\left [\frac{\sqrt{5}}{2}, \frac{-1}{\sqrt{5}} \right ]$
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
Here sum of zeroes $\frac{-3}{2\sqrt{5}}$
Product of zeroes $=-\frac{1}{2}$
Let p(x) is the required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeros)
$p(x)=x^{2}-\left (\frac{-3}{2\sqrt{5}} \right )x+ \frac{-1}{2}$
$p(x)=x^{2}+\frac{3}{2\sqrt{5}} x- \frac{1}{2}$
Multiplying by $2 \sqrt{5}$ we get
$p(x)=2 \sqrt{5}x^{2}+3x-\sqrt{5}$
Hence, $2 \sqrt{5}x^{2}+3x-\sqrt{5}$ is the required polynomial
$p(x)=2 \sqrt{5}x^{2}+3x-\sqrt{5}$
$=2 \sqrt{5}x^{2}+5x-2x -\sqrt{5}$
=$\sqrt{5}(2x+\sqrt{5})-1(2x+\sqrt{5})$
=$(\sqrt{5}x -1)(2x+\sqrt{5})$
$\left [\frac{-\sqrt{5}}{2}, \frac{1}{\sqrt{5}} \right ]$ are the zeroes of p(x).

Question:2

Solution. Here the given cubic polynomial is $x^3 -6x^2 + 3x + 10$
A = 1, B = –6, C = 3, D = 10
Given that, $\alpha = a, \beta = a + b, \gamma = a + 2b$
We know that, $\alpha +\beta +\gamma =\frac{-B }{A}$
$a + a + b + a + 2b =\frac{-(-6)}{1}$
$3a + 3b = 6$
3(a + b) = 6
a + b = 2 …..(1)
$\alpha \beta +\beta \gamma +\alpha = \frac{C}{A}$
$a(a + b) + (a + b) (a + 2b) + (a + 2b) (a) =\frac{3}{1}$
$a(2) + 2(a + 2b) + a^2 + 2ab = 3$ (Q a + b = 2)
$2a + 2a + 4b + a^2 + 2ab = 3$
$4a + 4b + a^2 + 2ab = 3$
$4a + 4(2 - a) + a^2 + 2a(2 - a) = 3$ (Q a + b = 2)
$4a + 8 - 4a + a^2 + 4a - 2a^2 = 3$
$a^2 + 4a - 2a^2 + 8 - 3 = 0$
$-a^2 + 4a + 5 = 0$
$a^2 - 4a - 5 = 0$
$a^2 - 5a + a - 5 = 0$
$a(a - 5)+1 (a - 5) = 0$
$(a -5) ( a + 1) = 0$
a = 5, –1
Put a = 5 in (1) put a = –1 in (1)
5 + b = 2 –1 + b = 2
b = –3 b = 3
Hence, value of a = 5, b = –3 and a = –1, b = 3
put a = 5, b = –3 and we get zeroes
a = 5
a + b = 5 – 3 = 2
a + 2b = 5 + 2(–3) = –1
Hence, zeroes are 5, 2, –1.

Question:3

Answer. $\left [\frac{-2\sqrt{2}}{3}, \frac{-1}{\sqrt{2}} \right ]$
Given cubic polynomial is $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$
If $\sqrt{2}$ is a zero of the polynomial then (x – $\sqrt{2}$ ) is a factor of $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$

$6x^3 + \sqrt{2} x^2 - 10x -4\sqrt{2} = (6x^2 + 7\sqrt{2} x + 4) (x -\sqrt{2} )$
$= (6x^2 + 4\sqrt{2} x + 3\sqrt{2} x + 4\sqrt{2}) (x -\sqrt{2} )$
$= (2x(3x + 2\sqrt{2} ) +\sqrt{2} (3x + 2\sqrt{2} )) (x -\sqrt{2} )$
$= (3x + 2 \sqrt{2}) (2x +\sqrt{2} ) (x -\sqrt{2} )$=0
$3x + 2\sqrt{2} = 0$ $2x + \sqrt{2} = 0$
$x=\frac{-2\sqrt{2}}{3}$ $x=\frac{-1}{\sqrt{2}}$
Hence, $\left [\frac{-2\sqrt{2}}{3}, \frac{-1}{\sqrt{2}} \right ]$ are the other two zeroes.

Question:4

The given polynomial is $2x^4 + x^3 - 14x^2 + 5x + 6$
Here, $x^2 + 2x + k$ is a factor of $2x^4 + x^3 - 14x^2 + 5x + 6$
Use division algorithm

Here, $x(7k + 21) + (6 + 8k + 2k^2)$ is remainder
It is given that $x^2 + 2x + k$ is a factor hence remainder = 0
$x(7k + 21) + (6 + 8k + 2k^2) = 0.x + 0$
By comparing L.H.S. and R.H.S.
7k + 21 = 0
k = – 3 ….(1)
$2k^2 + 8k + 6 = 0$
$2k^2 + 2k + 6k + 6 = 0$
2k(k + 1) + 6(k + 1) = 0
(k + 1) (2k + 6) = 0
k = –1, –3 …..(2)
From (1) and (2)
k = –3
Hence, $2x^4 + x^3 - 14x^2 + 5x + 6 = (x^2 + 2x + k) (2x^2 - 3x - 8 - 2k)$
Put k = –3
$= (x^2 + 2x - 3) (2x^2 - 3x - 8 + 6)$
$= (x^2 + 2x - 3) (2x^2 - 3x - 2)$
$= (x^2 + 3x - x - 3) (2x^2 - 4x + x - 2)$
$= (x(x + 3) - 1(x + 3)) (2x(x - 2) + 1(x - 2))$
$= (x + 3)(x -1)(x - 2) (2x + 1)$
x + 3 = 0 x – 1 = 0 x – 2 = 0 2x + 1 = 0
x = –3 x = 1 x = 2 2x = – 1
$x=\frac{-1}{2}$
Here zeroes of $x^2 + 2x - 3$is –3, 1
Zeroes of $2x^2 - 3x - 2 is 2, \frac{-1}{2} .$

Question:5

Answer. $\sqrt{5},\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$
Solution.
The given cubic polynomial is $x^3 -3 \sqrt{5}x^2 + 13x - 3\sqrt{5}$
Hence $x-\sqrt{5}$ is a factor
Use divided algorithm

$x^3 - 3\sqrt{3} x^2 + 13x - 3\sqrt{5}=(x-\sqrt{5}) (x^{2}-2\sqrt{5}+3)$
$=(x-\sqrt{5}) (x^{2}-2\sqrt{5}+3)$
$x^{2}-2\sqrt{5}+3$
a = 1, b = $-2\sqrt{5}$ , c = 3
$x=\frac{-b\pm \sqrt{b^{2}}-4ac}{2a}$
$x=\frac{2\sqrt{5}\pm \sqrt{20-12}}{2}$
$x=\frac{2\sqrt{5}\pm 2\sqrt{2}}{2}=\sqrt{5}\pm \sqrt{2}$
$x=\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$

Hence, $\sqrt{5},\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$ are the zeroes of the polynomial.

Question:6

Solution.
Given :-
$q(x) = x^3 + 2x^2 + a$, $p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$
Hence, q(x) is a factor of p(x) use divided algorithm

Since $x^3 + 2x^2 + a$ is a factor hence remainder = 0
$x^2(-1 - a) + 3x(1 + a) + (b - 2a) = 0.x^2 + 0.x + 0$
By comparing
– 1 – a = 0 b – 2a = 0
a = – 1 2a = b
put a = –1
2(–1) = b
b = –2
Hence, a = –1, b = –2
$x^5 - x^4 - 4x^3 + 3x^2 + 3x - 2 = (x^3 + 2x^2 - 1) (x^2 - 3x + 2)$
$= (x^3 + 2x^2- 1) (x^2 - 2x - x + 2)$
$= (x^3 + 2x^2 - 1) (x(x - 2) - 1(x - 2))$
$= (x^3 + 2x^2 -1) (x - 2) (x - 1)$
x – 2 = 0 x – 1 = 0
x = 2 x = 1
Hence, 2, 1 are the zeroes of p(x) which are not the zeroes of q(x).

## NCERT Exemplar Solutions Class 10 Maths Chapter 2 Important Topics:

• Remainder theorem and zeros of a polynomial.
• Division method of a polynomial is discussed in detail.
• The graphical representation of polynomials is discussed in this chapter.
• NCERT exemplar Class 10 Maths solutions chapter 2 discusses the coefficient of polynomial and their properties.
• The relation between zeros of the polynomial and coefficient of the polynomial is also discussed.

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## NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Real Numbers Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry & Its Equations Chapter 9 Circles Chapter 10 Constructions Chapter 11 Areas related to Circles Chapter 12 Surface Areas and Volumes Chapter 13 Statistics and Probability

## Features of NCERT Exemplar Class 10 Maths Solutions Chapter 2:

• These Class 10 Maths NCERT exemplar chapter 2 solutions provide a detailed knowledge of polynomials.

• The basics of this chapter are covered in Class 9. If any student wants to pursue mathematics in higher classes and appear for competitive exams like JEE Advanced, this chapter holds great importance.

• All sets of problems related to Class 10 Polynomials can be studied and practiced through these Class 10 Maths NCERT exemplar solutions chapter 2 Polynomials and are sufficient to solve books such as RS Aggarwal Class 10 Maths, RD Sharma Class 10 Maths, NCERT Class 10 Maths.

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### Check Chapter-Wise Solutions of Book Questions

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

### Must, Read NCERT Solution Subject Wise

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

### Also, Check NCERT Books and NCERT Syllabus here

1. Is the chapter Polynomials important for Board examinations?

The chapter of Polynomials is important for Board examinations as it holds around 5-6% weightage of the whole paper.

2. Is the chapter Polynomials important for competitive examinations like JEE Advanced?

The chapter on Polynomials serves as a key topic for JEE Advanced and clarity of concepts beginning from lower classes such as Class 10 can help the students to build a strong foundation for competitive exams as well as higher classes.

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Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9