NCERT Exemplar Class 10 Maths Solutions Chapter 2 Polynomials

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) $\frac{-2}{3}$

Answer. [A]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain 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)
$p(-3) = (k - 1) (-3)^2 + k(-3) + 1$
$\Rightarrow0 = (k - 1) (9) -3k + 1$
$\Rightarrow0 = 9k - 9 - 3k + 1$
$\Rightarrow0 = 6k - 8$
$\Rightarrow 6k = 8$
$\Rightarrow k =\frac{8}{6}$
$\Rightarrow k =\frac{4}{3}$

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) $2x^2 + 2x -24$

Answer. [C]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain 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 and put x = 4
$p(-3) = (-3)^2 - (-3) + 12$

$= 9 + 3 + 12 = 24 \neq 0$

$p(4) = (4)^2 - 4 + 12$
$= 16 - 4 + 12 = 24 \neq 0$
(B) $p(x) = x^2 + x + 12$
put x = –3 and put x = 4
$p(-3) = (-3)^2 - 3 + 12$

$= 9 + 9 = 18 \neq 0$

$p(4) = (4)^2 + 4 + 12$
$= 16 + 16 = 32 \neq$ 0
(C) $p(x)=\frac{x^2}{2} - \frac{x}{2} -6$
put x = –3 and put x = 4
$p(-3)=\frac{(-3)^2}{2} - \frac{-3}{2} -6$

$=\frac{9}{2}+\frac{3}{2}-6$

$=\frac{9+3-12}{2}=0$

$p(4)=\frac{(4)^2}{2} - \frac{4}{2} -6$
$=\frac{16}{2}-2-6$
$= 8 - 8 = 0$
(D) $p(x) = 2x^2 + 2x - 24$
put x = –3 and put x = 4
$p(-3) = 2(-3)^2 + 2(-3) - 24$

$= 18 -6 - 24$

$= -12 \neq 0$

$p(4) = 2(4)^2 + 2(4) - 24$
$= 32 + 8 - 24$
$= 16 \neq 0$
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

Answer. [D]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a 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.
$p(2) = (2)^2 + (a + 1) (2) + b$
$\Rightarrow 0 = 4 + 2a + 2 + b$
$\Rightarrow 0 = 6 + 2a + b$
$\Rightarrow 2a + b = -6$ ….(1)
$p(-3) = (-3)^2 + (a + 1) (-3) + b$
$\Rightarrow 0 = 9 - 3a - 3 + b$
$\Rightarrow 3a - b = 6$ ….(2)
Add equations (1) and (2)
$2a + b + 3a - b = -6 + 6$
$\Rightarrow 2a + 3a = 0$
$\Rightarrow 5a = 0$
$\Rightarrow a = 0$
put a = 0 in (1)
2(0) + b = –6
$\Rightarrow b = –6$
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

Answer. [D]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Let the polynomial is $ax^2 + bx + c = 0$ ….(*)
We know that sum of zeroes $-\frac{b}{a}$
$- 2 + 5= -\frac{b}{a}$
$\Rightarrow \frac{3}{1}=-\frac{b}{a}$ …..(1)
Multiplication of zeroes $=\frac{c}{a}$
$-2 \times 5=\frac{c}{a}$
$\Rightarrow -\frac{10}{1}=\frac{c}{a}$ …..(2)
From equations (1) and (2), it is clear that
a = 1, b = –3, c = –10
Put values of a, b, and c in equation (*)
$x^3 - 3x - 10 = 0$ ….(3)
But we can multiply or divide equations. (3) by any real number except 0, and the zeroes remain the same.
Hence, there are an infinite number of polynomials that 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) $-\frac{b}{a}$

Answer. [B]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a 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 the 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

Answer. [A]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s) and a quadratic polynomial is a 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 $p(x)=x^3 + ax^2 + bx + c$
$p(-1) = (-1)^3 + a(-1)^2 + b(-1) + c$
$\Rightarrow 0 = -1 + a - b + c ( Q=-1 is zero)$
$\Rightarrow c = 1 - a + b$
We know that
$\alpha \beta \gamma =\frac{-d}{a}$
Here, a = 1, b = a, c = b, d = c
So, $\alpha \beta \gamma =\frac{-c}{1}$
$\Rightarrow \beta \gamma =\frac{-(1-a+b)}{\alpha }$
$\Rightarrow \beta \gamma =\frac{-(1-a+b)}{-1 } = 1 -a + b$ ( using equation (1))
Hence, the product of the 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

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

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

$\Rightarrow x = -97.7$
The value of both the zeroes is 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

Answer. [A]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a polynomial of degree 2.
Here, the given quadratic polynomial is $x^2 + kx + k$
$a = 1, b = k, c = k$
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\Rightarrow x=\frac{-k\pm \sqrt{k^{2}-4k}}{2}$
$\Rightarrow x=\frac{-k\pm \sqrt{k(k-4)}}{2}$
$k(k - 4)$ must be greater than 0
Hence, the value of k is either less than 0 or greater than 4.
If the value of k is less than 0, only one zero is positive.
If the 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

Answer. [C]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a polynomial of degree 2.
Here the given polynomial is $ax^2 + bx + c, c \neq 0$
$a = a, b = b, c = c$
If both the zeroes are equal, then
$b^2 - 4ac = 0$
$\Rightarrow b^2 =4ac$ ….(1)
(A) c and a have opposite signs
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 the correct one.
(C) c and a have the 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.

Answer. [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 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 the 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 $= (x_1)(x_2) = \frac{c}{a}$
$\Rightarrow (x_1)(-x_1) = \frac{c}{a}$
$\Rightarrow -x_{1}^{2} =b$ ( because c= b, a = 1)
Put the 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

Which of the following is not the graph of a quadratic polynomial?

Answer:

Answer. [D]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
If p(x) is a 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

(i)Answer the following and justify:
Can $x^2 - 1$ be the quotient on division of $x^6 + 2x^3 + x - 1$ by a polynomial in x of degree 5?
(ii)Answer the following and justify:
What will the quotient and remainder be on division of $ax^2 + bx + c \ by \ px^3 + qx^2 + rx+ s, p\neq 0?$$ax^2 + bx + c \ by \ px^3 + qx^2 + rx+ s, p\neq 0?$
(iii)Answer the following and justify:
If on division of a polynomial p (x) by a polynomial g (x), the quotients zero, what is the relation between the degrees of p (x) and g (x)?
(iv) Answer the following and justify:
If on division of a non-zero polynomial p (x) by a polynomial g (x), the remainder is zero, what is the relation between the degrees of p (x) and g (x)?
(v) Answer the following and justify:
Can the quadratic polynomial $x^2 + kx + k$have equal zeroes for some odd integer k > 1?

Answer:

(i) Answer. [false]
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 $p(x)=g(x) \times q(x)+r(x)$

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 $p(x)=g(x) \times q(x)+r(x)$

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)
Answer. [false]
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
$\therefore d = (k)^2 - 4 \times 1 \times k = 0$
$k^2 - 4k = 0$
$\Rightarrow k(k -4) = 0$
$\Rightarrow$ 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, the given statement is not correct.

Question:2

Are the following statements ‘True’ or ‘False’? Justify your answers.
(i)If the zeroes of a quadratic polynomial $ax^2 + bx + c$ are both positive, then a, b and c all have the same sign
(ii) If the graph of a polynomial intersects the x-axis at only one point, it cannot be a quadratic polynomial.
(iii) If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.
(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.
(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.
(vi) If all three zeroes of a cubic polynomial $x^3 + ax^2 - bx + c$ are positive, then at least one of a, b and c is non-negative.
(vii) The only value of k for which the quadratic polynomial $kx^2 + x + k$ has equal zeros is$\frac{1}{2}$

(i) Answer. [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.
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.
(ii) Answer. [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)
$\Rightarrow (x + 2)^2 = 0$

$\Rightarrow x + 2 = 0$
$\Rightarrow x =-2$
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)
Answer. [True]
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)
Answer. [True]
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 $p(y) = (y - \alpha _1)(y - \alpha _2)(y -\alpha _3)$
$\Rightarrow p(y) = (y - 0) (y - 0) (y - \alpha _3)$
$\Rightarrow p(y) = y^3 -y^2\alpha _3$
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)
Answer. [True]
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 $=\frac{\text{coefficient of} ~x^2}{\text{coefficient of} ~x^3}$
$\alpha +\beta +\gamma =-\frac{b}{a}$
It is given that zeroes are negative then $\alpha = -\alpha , \beta = -\beta$, and $\gamma = -\gamma$
$\Rightarrow -\alpha +(-\beta) +(-\gamma) =-\frac{b}{a}$
$\Rightarrow-\alpha -\beta -\gamma =-\frac{b}{a}$
$\Rightarrow-\left (\alpha +\beta +\gamma \right ) =-\frac{b}{a}$
$\Rightarrow \left (\alpha +\beta +\gamma \right ) =\frac{b}{a}$ …….(1)
That is $\frac{b}{a}>0$
Sum of the products of two zeroes at a time $=\frac{\text{coefficient of} ~x}{\text{coefficient of} ~x^3}$
$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
Replace $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$
$(-\alpha ) (-\beta) +(-\beta) (-\gamma) +(-\gamma) (-\alpha )=\frac{c}{a}$
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$ …..(2)
That is $\frac{c}{a}>0$
Product of all zeroes $=\frac{\text{-constant terms}}{\text{coefficient of} ~x^3}$
$\alpha \beta \gamma =-\frac{d}{a}$
Replace $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$
$(-\alpha) (-\beta) (-\gamma) =-\frac{d}{a}$
$\Rightarrow -(\alpha \beta \gamma)=-\frac{d}{a}$
$\Rightarrow \alpha \beta \gamma =\frac{d}{a}$ ……(3)
That is $\frac{d}{a}>0$
From equations (1), (2) and (3) we conclude that all the coefficient and the constant term of the polynomial have the same sign.
Hence, the given statement is true.
(vi)
Answer. [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).
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 $=\frac{\text{coefficient of} ~x^2}{\text{coefficient of} ~x^3}$
$\alpha +\beta +\gamma =-\frac{a}{1}=-a$
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 $=\frac{\text{coefficient of} ~x}{\text{coefficient of} ~x^3}$
$\alpha \beta +\beta \gamma +\gamma \alpha =-\frac{b}{1}=-b$
Also, here b is negative
Product of all zeroes $=\frac{\text{-constant terms}}{\text{coefficient of} ~x^3}$
$\alpha \beta \gamma =\frac{-c}{1}=-c$
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)
Answer. [False]
Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Quadratic polynomial: when the degree of polynomial is two then the polynomial is called quadratic polynomial.
Let $p(x)=kx^2 + x + k$
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
$b^2 - 4ac = 0$
$p(x)=kx^2 + x + k$
Here, a = k, b = 1, c = k
$b^2 - 4ac = 0$
$\Rightarrow (1)2 -4 \times k \times k = 0$
$\Rightarrow 1 - 4k^2 = 0$
$\Rightarrow 1 = 4k^2$
$\Rightarrow \frac{1 }{4}= k^2$
$\Rightarrow k=\pm \frac{1}{2}$
$\therefore$ When $+\frac{1}{2}$ and $-\frac{1}{2}$ then the given quadratic polynomial has equal zeroes.

Question:1

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$4x^2 - 3x - 1$

Answer. $\left [1, -\frac{1}{4} \right ]$
Zeroes : zeroes of polynomial are the value(s) that makes it equal to 0.
$4x^2 - 3x - 1=0$
$\Rightarrow 4x^2 - 4x + x - 1=0$
$\Rightarrow 4x(x - 1) + 1(x - 1)=0$
$\Rightarrow (x -1) (4x + 1)$
x – 1 = 0 or 4x + 1 = 0
x = 1 or 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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$3x^2 + 4x - 4$

Answer. $\left [-2, \frac{2}{3} \right ]$
Solution. Zeroes: zeroes of the polynomial are the value(s) that make it equal to 0.
$3x^2 + 4x - 4=0$
$\Rightarrow 3x^2 + 6x - 2x -4=0$
$\Rightarrow 3x(x + 2) - 2(x + 2)=0$
$\Rightarrow (x + 2) (3x - 2)=0$
$x + 2 = 0$ or $3x - 2 = 0$
$x = -2$ or $x =\frac{2}{3}$
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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$5t^2 + 12t + 7$

Answer. $\left [-1, -\frac{7}{5} \right ]$
Zeroes : zeroes of the polynomial are the values(s) the makes it equal to 0
$5t^2 + 12t + 7=0$
$\Rightarrow 5t^2 + 5t + 7t + 7=0$
$\Rightarrow 5t(t + 1) + 7(t + 1)=0$
$\Rightarrow (t + 1) (5t + 7)=0$
t + 1 = 0 or 5t + 7 = 0
t = –1 or $t =\frac{-7}{5}$
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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$t^3 - 2t^2 - 15t$

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)
$\Rightarrow t(t^2 - 2t - 15)$ =0
$\Rightarrow t(t^2 - 5t + 3t - 15)$=0
$\Rightarrow t(t(t - 5) + 3(t - 5))$=0
$\Rightarrow t(t - 5) (t + 3)$=0
t = 0 or t – 5 = 0 or t + 3 = 0
t = 5 or 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$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$
$(0 \times 5) + (5 \times - 3) + (-3 \times 0) = -\frac{15}{1}$
$\Rightarrow -15 = -15$
Hence proved.

Question:5

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$2x^{2}+\frac{7}{2}x+\frac{3}{4}$

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
$\Rightarrow 8x^2 + 12x + 2x + 3$ =0
$\Rightarrow 4x(2x + 3) + 1(2x + 3)$=0
$\Rightarrow (2x + 3) (4x + 1)$ =0
$\Rightarrow 2x + 3 = 0$ or $4x + 1 = 0$
$\Rightarrow x=\frac{-3}{2}$ or $x=\frac{-1}{4}$
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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$4x^{2}+5 \sqrt{2}x-3$

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
$\Rightarrow 4x^{2}+6 \sqrt{2}x-\sqrt{2}x-3=0$
$\Rightarrow 2\sqrt{2}x (\sqrt{2}x+3)-1 (\sqrt{2}x+3)=0$
$\Rightarrow (\sqrt{2}x+3)(1-2\sqrt{2}x)=0$
$\Rightarrow (\sqrt{2}x+3) =0$ $(1-2\sqrt{2}x) =0$
$\Rightarrow x=\frac{-3}{\sqrt{2}}$ or $x=\frac{1}{2\sqrt{2}}$
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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$2s^{2}-(1+2\sqrt{2})s+\sqrt{2}$

Answer. $\left [\frac{1}{2}, \sqrt{2} \right ]$
Solution. Zeroes: zeroes of the polynomial are the value(s) that make it equal to 0.
$2s^{2}-(1+2\sqrt{2})s+\sqrt{2}$ = 0
$\Rightarrow 2s^{2}-s-2\sqrt{2}s+\sqrt{2}$=0
$\Rightarrow s(2s-1)-\sqrt{2}(2s-1)$=0
$\Rightarrow (s-\sqrt{2})(2s-1)$=0
$\Rightarrow 2s-1$=0 or $(s-\sqrt{2})=0$
$\Rightarrow s=\frac{1}{2}$ or $s=\sqrt{2}$
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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$v^{2}+4\sqrt{3}v-15$

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
$\Rightarrow v^{2}+5\sqrt{3}v-\sqrt{3}v-15$=0
$\Rightarrow v(v+ 5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})$=0
$\Rightarrow (v+ 5\sqrt{3})(v-\sqrt{3})$=0
$\Rightarrow (v+ 5\sqrt{3}) =0$ or $(v-\sqrt{3})=0$
$\Rightarrow v =-5\sqrt{3}$ or $v =\sqrt{3}$
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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$y^{2}+\frac{3}{2}\sqrt{5}y-5$

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
$\Rightarrow 2y^{2}+4\sqrt{5}y-\sqrt{5}y-10$=0
$\Rightarrow 2y(y + 2 \sqrt{5}y)-\sqrt{5}(y + 2 \sqrt{5}y)$=0
$\Rightarrow (y + 2 \sqrt{5}y)(2y-\sqrt{5})$=0
$\Rightarrow (y + 2 \sqrt{5}) =0$ or $(2y-\sqrt{5})=0$
$\Rightarrow y=-2\sqrt{5}$ or $y=\frac{\sqrt{5}}{2}$
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

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$7y^{2}-\frac{11}{3}y-\frac{2}{3}$

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
$\Rightarrow 21y^2 - 14y + 3y - 2$=0
$\Rightarrow 7y(3y -2) + 1 (3y - 2)$=0
$\Rightarrow (3y - 2) (7y + 1)$=0
$\Rightarrow 3y - 2 = 0$ or $7y + 1 = 0$
$\Rightarrow y = \frac{2}{3}$ or $y = \frac{-1}{7}$
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)
$\Rightarrow p(x)=x^{2}-\left (\frac{-3}{2\sqrt{5}} \right )x+ \frac{-1}{2}$
$\Rightarrow p(x)=x^{2}+\frac{3}{2\sqrt{5}} x- \frac{1}{2}$
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

Given that the zeroes of the cubic polynomial $x^3 -6x^2 + 3x + 10$ are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Answer. [5, 2, –1]
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, $\alpha +\beta +\gamma =\frac{-B }{A}$
$a + a + b + a + 2b =\frac{-(-6)}{1}$
$\Rightarrow 3a + 3b = 6$
$\Rightarrow$3(a + b) = 6
$\Rightarrow$a + b = 2 …..(1)
$\alpha \beta +\beta \gamma +\alpha = \frac{C}{A}$
$\Rightarrow a(a + b) + (a + b) (a + 2b) + (a + 2b) (a) =\frac{3}{1}$
$\Rightarrow a(2) + 2(a + 2b) + a^2 + 2ab = 3$ (Q a + b = 2)
$\Rightarrow 2a + 2a + 4b + a^2 + 2ab = 3$
$\Rightarrow 4a + 4b + a^2 + 2ab = 3$
$\Rightarrow 4a + 4(2 - a) + a^2 + 2a(2 - a) = 3$ (Q a + b = 2)
$\Rightarrow 4a + 8 - 4a + a^2 + 4a - 2a^2 = 3$
$\Rightarrow a^2 + 4a - 2a^2 + 8 - 3 = 0$
$\Rightarrow -a^2 + 4a + 5 = 0$
$\Rightarrow a^2 - 4a - 5 = 0$
$\Rightarrow a^2 - 5a + a - 5 = 0$
$\Rightarrow a(a - 5)+1 (a - 5) = 0$
$\Rightarrow (a -5) ( a + 1) = 0$
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

Given that $\sqrt{2}$ is a zero of the cubic polynomial $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$ , find its other two zeroes.

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

Find k so that $x^2 + 2x + k$ is a factor of $2x^4 + x^3 - 14x^2 + 5x + 6$. Also find all the zeroes of the two polynomials.

Answer:

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)
$2k^2 + 8k + 6 = 0$
$\Rightarrow 2k^2 + 2k + 6k + 6 = 0$
$\Rightarrow$2k(k + 1) + 6(k + 1) = 0
$\Rightarrow$(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

Given that $x-\sqrt{5}$ is a factor of the cubic polynomial $x^3 -3 \sqrt{5}x^2 + 13x - 3\sqrt{5}$ ,find all the zeroes of the polynomial.

Answer:

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
$x=\frac{-b\pm \sqrt{b^{2}}-4ac}{2a}$
$\Rightarrow x=\frac{2\sqrt{5}\pm \sqrt{20-12}}{2}$
$\Rightarrow x=\frac{2\sqrt{5}\pm 2\sqrt{2}}{2}=\sqrt{5}\pm \sqrt{2}$
$\Rightarrow x=\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$

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

Question:6

For which values of a and b, are the zeroes of $q(x) = x^3 + 2x^2 + a$ also the zeroes of the polynomial $p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$? Which zeroes of p(x) are not the zeroes of q(x)?

Answer: [2, 1]
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).