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#### Question 1:

Choose the correct answer from the given four options:
If one of the zeroes of the quadratic polynomial (k – 1) x2 + kx + 1 is –3, then the value of k is
(A) $\frac{4}{3}$

(B) $\frac{-4}{3}$

(C) $\frac{2}{3}$

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

Let α and β be the zeroes of the given polynomial and α −3
We know that
$\alpha +\beta =\frac{-b}{a}$

Putting the value of  β in (1) we get,
$-3+\frac{-1}{3\left(k-1\right)}=\frac{-k}{k-1}\phantom{\rule{0ex}{0ex}}\frac{k}{k-1}+\frac{-1}{3\left(k-1\right)}=3\phantom{\rule{0ex}{0ex}}\frac{3k-1}{3\left(k-1\right)}=3\phantom{\rule{0ex}{0ex}}3k-1=9k-9\phantom{\rule{0ex}{0ex}}8=6k\phantom{\rule{0ex}{0ex}}k=\frac{4}{3}$
Hence, the correct answer is option (A).

#### Question 2:

Choose the correct answer from the given four options:
A quadratic polynomial, whose zeroes are –3 and 4, is
(A) x2x + 12

(B) x2 + x + 12

(C) $\frac{{x}^{2}}{2}-\frac{x}{2}-6$

(D) 2x2 + 2x – 24

Let α and β be the zeroes of a quadratic polynomial.
Then quadratic polynomial is $k\left[{x}^{2}-\left(\alpha +\beta \right)x+\alpha \beta \right]$.
So, a quadratic polynomial with zeroes −3 and 4 is

Hence, the correct answer is option (C).

#### Question 3:

Choose the correct answer from the given four options:
If the zeroes of the quadratic polynomial x2 + (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

Let and be the zeroes of polynomial $A{x}^{2}+Bx+C$.
Then we know that
$\alpha +\beta =\frac{-B}{A}$   .....(i)

and $\alpha \beta =\frac{C}{A}$    .....(ii)
Given 2 and −3 are the zeroes of the quadratic polynomial x2 + (a + 1) x + b,

using (i)
$2+\left(-3\right)=\frac{-\left(a+1\right)}{1}\phantom{\rule{0ex}{0ex}}-1=-a-1\phantom{\rule{0ex}{0ex}}a=0$
using (ii)
$2×\left(-3\right)=\frac{b}{1}\phantom{\rule{0ex}{0ex}}b=-6$

Hence, the correct answer is option (D)

#### Question 4:

Choose the correct answer from the given four options:
The number of polynomials having zeroes as –2 and 5 is
(A) 1
(B) 2
(C) 3
(D) more than 3

Quadratic polynomial having zeroes $\alpha$ and $\beta$ is given by $k\left[{x}^{2}-\left(\alpha +\beta \right)x+\alpha \beta \right]$.
So, the quadratic polynomial having zeroes −2 and 5 is given by
$k\left[{x}^{2}-\left(-2+5\right)x+\left(-2\right)×5\right]\phantom{\rule{0ex}{0ex}}=k\left[{x}^{2}-3x-10\right]$
where k is an arbitrary constant.
So, by putting k = 1, 2, 3, 4....., we can have infinitely many polynomials.

Hence, the correct answer is option (D).

#### Question 5:

Choose the correct answer from the given four options:
Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is
(A) $-\frac{c}{a}$

(B) $\frac{c}{a}$

(C) 0

(D) $-\frac{b}{a}$

Let  be the zeroes of the cubic polynomial ax3 + bx2 + cx d.
We know that,
$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}\phantom{\rule{0ex}{0ex}}$
Let $\alpha =0$
$⇒0×\beta +\beta ×\gamma +\gamma ×0=\frac{c}{a}\phantom{\rule{0ex}{0ex}}⇒\beta \gamma =\frac{c}{a}$

Hence, the correct answer is option (B).

#### Question 6:

Choose the correct answer from the given four options:
If one of the zeroes of the cubic polynomial x3 + ax2 + 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

Let  be the zeroes of the cubic polynomial Ax3 + Bx2 + Cx + D.
We know that

and

Hence, the correct answer is option (A)

#### Question 7:

Choose the correct answer from the given four options:
The zeroes of the quadratic polynomial x2 + 99x + 127 are
(A) both positive
(B) both negative
(C) one positive and one negative
(D) both equal

To check the zeroes of a quadratic polynomial $a{x}^{2}+bx+c$, we have $D={b}^{2}-4ac$.
For x2 + 99x + 127 we have,
$\phantom{\rule{0ex}{0ex}}⇒D={\left(99\right)}^{2}-4×1×127\phantom{\rule{0ex}{0ex}}⇒D=9801-508\phantom{\rule{0ex}{0ex}}⇒D=9293>0\phantom{\rule{0ex}{0ex}}$
Therefore, both the zeroes will be real and distinct.
Since, a, b, c are all positive therefore, both zeroes will be negative.

Hence, the correct answer is option (B)

#### Question 8:

Choose the correct answer from the given four options:
The zeroes of the quadratic polynomial x2 + kx + k, k ≠ 0,
(A) cannot both be positive
(B) cannot both be negative
(C) are always unequal
(D) are always equal

To check the nature of the zeroes of a quadratic polynomial $a{x}^{2}+bx+c$, we have $D={b}^{2}-4ac$.
For quadratic polynomial x2 + kx k, we have .
$⇒D={k}^{2}-4k\phantom{\rule{0ex}{0ex}}⇒D=k\left(k-4\right)\phantom{\rule{0ex}{0ex}}$
Given $k\ne 0$,
$⇒$
roots will be equal
$⇒$no real roots
$⇒$roots are real i.e, roots can be positive or negative

So, in any case zeroes of the given quadratic polynomial cannot be both positive.
Hence, the correct answer is option (A).

#### Question 9:

Choose the correct answer from the given four options:
If the zeroes of the quadratic polynomial ax2 + bx + c, c ≠ 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

Given zeroes of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal.
$⇒D=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}$
b2 is always positive i.e, ${b}^{2}\ge 0$            (Since, square of any number is always greater than or equal to zero)
(i) When a and c are of opposite sign,
$ac<0\phantom{\rule{0ex}{0ex}}⇒4ac<0\phantom{\rule{0ex}{0ex}}⇒-4ac>0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-4ac>0$

(ii) When a and c are of same sign,
$ac>0\phantom{\rule{0ex}{0ex}}⇒4ac>0\phantom{\rule{0ex}{0ex}}⇒-4ac<0\phantom{\rule{0ex}{0ex}}$
$⇒{b}^{2}\ge 0$

So, ${b}^{2}-4ac=0$ is possible for some values of a, b and c when a and c are of same sign.

Hence, the correct answer is option (C).

#### Question 10:

Choose the correct answer from the given four options
If one of the zeroes of a quadratic polynomial of the form x2 + 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.

Let  be the zeroes of quadratic polynomial x2 + ax b.
Given one of the zero is negative of other i.e, $\beta =-\alpha$, then
$\alpha +\beta =-a$                   [Since, sum of the zeroes = ]

$\phantom{\rule{0ex}{0ex}}⇒\alpha -\alpha =-a\phantom{\rule{0ex}{0ex}}⇒a=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\alpha \beta =b\phantom{\rule{0ex}{0ex}}$                               [Since, product of zeroes = ]
$⇒\alpha ×\left(-\alpha \right)=b\phantom{\rule{0ex}{0ex}}⇒b=-{\alpha }^{2}$

So, the quadratic polynomial becomes ${x}^{2}+0x-{\alpha }^{2}$.

Hence, the correct answer is option (A).

#### Question 11:

Choose the correct answer from the given four options.
Which of the following is not the graph of a quadratic polynomial?
(A) (B) (C) (D) For quadratic polynomial ax2 + bx + c, a ≠ 0, the curve of the quadratic polynomial crosses the X-axis on at most two points.
But, in option (D) the curve crosses the X-axis on three points, so it does not represent quadratic polynomial.

Hence, the correct answer is option (D).

#### Question 1:

(i) Can x2 – 1 be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5?
(ii) What will the quotient and remainder be on division of ax2 + bx + c by px3 + qx2 + rx + s, p ≠ 0?
(iii) If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p(x) and g(x)?
(iv) 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) Can the quadratic polynomial x2 + kx + k have equal zeroes for some odd integer k > 1?

(i) By Euclid's division algorithm,
p(x) = g(x) × q(x) + r(x)
Here,
p(x) = x6 + 2x3 + x – 1
g(x) = polynomial of degree 5
q(x) = x2 – 1
Let the degree of r(x) be y.
So,
Deg(6) = Deg(5) × Deg(2) + Deg[r(x)]
⇒ Deg(6) = Deg(5 + 2) + Deg[r(x)]
⇒ 6 = 5 + 2 + y    .....(1)
For (1) to be possible, the value of can only be 1.

Hence, x2 – 1 cannot be the quotient on division of x6 + 2x3 + x – 1 by a polynomial in x of degree 5.

(ii) By Euclid's division algorithm,
p(x) = g(x) × q(x) + r(x)
Here,
p(x) = px3 + qx2 + rx s
g(x) = ax2 + bx c

The degree of dividend is always greater than the divisor but here, deg[p(x)] < deg[g(x)] therefore, quotient will be 0 and remainder will be p(x).

(iii) When the quotient is 0, the deg[p(x)] < deg[g(x)].

(iv) When the remainder is 0, g(x) is a factor of p(x) and deg[g(x)] ≤ deg[p(x)].

(v) To check the nature of zeroes of a quadratic polynomial $a{x}^{2}+bx+c$ we have, $D={b}^{2}-4ac$.
For x2 + kx k,

$⇒D={k}^{2}-4k\phantom{\rule{0ex}{0ex}}⇒D=k\left(k-4\right)$
For equal zeroes, $D=0$, which is only possible when.
But it is given that k is an odd integer greater than 1.

So, x2 + kx cannot have equal zeroes for odd integer  > 1.

#### Question 2:

(i) If the zeroes of a quadratic polynomial ax2 + 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 x3 + ax2bx + 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 kx2 + x + k has equal zeros is $\frac{1}{2}$.

(i) Let  be the positive roots of the polynomial.
Then,
$\alpha +\beta =\frac{-b}{a}$
Here, LHS is positive but RHS is negative.
⇒ Either b or have to be negative.
So, b and have opposite signs.

Hence, the statement is false.

(ii) Graph of a quadratic polynomial with equal roots intersect the x-axis at only one point.

Hence, the given statement is false.

(iii) Graph of a polynomial of degree 3 having two equal roots and one other root will intersect the x-axis at 2 points.

Hence, the given statement is true.

(iv) Let the roots of the cubic polynomial be . Then,
f(x) = $\left(x-\alpha \right)\left(x-\beta \right)\left(x-\gamma \right)$
⇒ f(x) = $\left(x-\alpha \right)\left(x-0\right)\left(x-0\right)$
⇒ f(x) = $x×x×\left(x-\alpha \right)$
⇒ f(x) = ${x}^{3}-\alpha {x}^{2}$

So, there is no linear and constant terms.

Hence, the given statement is true.

(v) Let  be the negative roots of the cubic polynomial.
Then,

So, all the coefficients and the constant term of the polynomial have the same sign.

Hence, the given statement is true.

(vi) Let  be the positive roots of the cubic polynomial.
Then,

Now, if a is positive, then b is negative, c is positive and d is negative.
So, at least one of a, b and c is negative.

Hence, the given statement is false.

(vii) To check the nature of zeroes of a quadratic polynomial $a{x}^{2}+bx+c$ we have, $D={b}^{2}-4ac$.
For kx2 + x + k

$⇒D=1-4{k}^{2}\phantom{\rule{0ex}{0ex}}$
For equal roots,

Hence, the given statement is false.

#### Question 1:

Find the zeroes of the following polynomial by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
4x2 – 3x – 1

4x2 – 3x – 1
$=4{x}^{2}-4x+x-1\phantom{\rule{0ex}{0ex}}-4x\left(x-1\right)+1\left(x-1\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(4x+1\right)\phantom{\rule{0ex}{0ex}}$

To find zeroes,

$⇒$zeroes of quadratic polynomial 4x2 – 3x – 1 are 1 or $\frac{-1}{4}$.
Now to check the relationship between the zeroes and the coefficients of the polynomial 4x2 – 3x – 1, we have .
Sum of zeroes = $1-\frac{1}{4}=\frac{3}{4}=\frac{-b}{a}$
Product of zeroes = $1×\left(\frac{-1}{4}\right)=\frac{-1}{4}=\frac{c}{a}$

Hence, the relation is verified.

#### Question 2:

Find the zeroes of the following polynomial by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
3x2 + 4x – 4

3x2 + 4x – 4
$=3{x}^{2}+6x-2x-4\phantom{\rule{0ex}{0ex}}=3x\left(x+2\right)-2\left(x+2\right)\phantom{\rule{0ex}{0ex}}=\left(3x-2\right)\left(x+2\right)$

To find zeroes

Zeroes of quadratic polynomial 3x2 + 4x – 4 are .
Now to verify the relationship between zeroes & coefficient of polynomial 3x2 + 4x – 4, we have
Sum of zeroes = $\frac{2}{3}+\left(-2\right)=\frac{-4}{3}=\frac{-b}{a}$
Product of zeroes = $\frac{2}{3}×\left(-2\right)=\frac{-4}{3}=\frac{c}{a}$

Hence, the relation is verified.

#### Question 3:

Find the zeroes of the following polynomial by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
5t2 + 12t + 7

5t2 + 12+ 7
$5{t}^{2}+5t+7t+7\phantom{\rule{0ex}{0ex}}=5t\left(t+1\right)+7\left(t+1\right)\phantom{\rule{0ex}{0ex}}=\left(5t+7\right)\left(t+1\right)\phantom{\rule{0ex}{0ex}}$
To find zeroes:

Zeroes of quadratic polynomial 5t2 + 12+ 7 are .
Now to find relationship between zeroes & coefficient of polynomial 5t2 + 12+ 7 we have,
Sum of zeroes = $\left(\frac{-7}{5}\right)+\left(-1\right)=\frac{-7}{5}-1=\frac{-12}{5}=\frac{-b}{a}$
Product of zeroes = $\left(\frac{-7}{5}\right)×\left(-1\right)=\frac{7}{5}=\frac{c}{a}$

Hence, the relation is verified.

#### Question 4:

Find the zeroes of the following polynomial by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
t3 – 2t2 – 15t

t3 – 2t2 – 15t
$t\left({t}^{2}-2t-15\right)\phantom{\rule{0ex}{0ex}}=t\left({t}^{2}-5t+3t-15\right)\phantom{\rule{0ex}{0ex}}=t\left[t\left(t-5\right)+3\left(t-5\right)\right]\phantom{\rule{0ex}{0ex}}=t\left(t-5\right)\left(t+3\right)$
To find the zeroes of given polynomial

$⇒$Zeroes of given polynomial are 0, 5 and −3.
Now to verify the relations between the zeroes and the coefficients of the polynomial t3 – 2t2 – 15we have,
Sum of zeroes = $0+5+\left(-3\right)=2=\frac{-b}{a}$
Sum of the product of two zeroes at a time = $0×5+5×\left(-3\right)+\left(-3\right)×0=-15=\frac{c}{a}$
Product of zeroes = $0×5×\left(-3\right)=0=\frac{-d}{a}$

Hence, the relation is verified.

#### Question 5:

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

$2{x}^{2}+\frac{7}{2}x+\frac{3}{4}$
$=\frac{1}{4}\left(8{x}^{2}+14x+3\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left(8{x}^{2}+12x+2x+3\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[4x\left(2x+3\right)+1\left(2x+3\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left(4x+1\right)\left(2x+3\right)$
To find the zeroes,
$2{x}^{2}+\frac{7}{2}x+\frac{3}{4}$ = 0

$⇒$Zeroes of the given polynomial are .
Now to verify the relation between the zeroes and the coefficients of the polynomial $2{x}^{2}+\frac{7}{2}x+\frac{3}{4}$ we have, .
Sum of the zeroes =
Product of zeroes = $\left(\frac{-1}{4}\right)×\left(\frac{-3}{2}\right)=\frac{3}{8}=\frac{c}{a}$

Hence, the relation is verified.

#### Question 6:

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

$4{x}^{2}+5\sqrt{2}x-3$
$=4{x}^{2}+6\sqrt{2}x-\sqrt{2}x-3\phantom{\rule{0ex}{0ex}}=2\sqrt{2}x\left(\sqrt{2}x+3\right)-1\left(\sqrt{2}x+3\right)\phantom{\rule{0ex}{0ex}}=\left(\sqrt{2}x+3\right)\left(2\sqrt{2}x-1\right)$
To find the zeroes

$⇒$Zeroes of the given polynomial are .
Now to verify the relation between the zeroes and the coefficients of the polynomial $4{x}^{2}+5\sqrt{2}x-3$ we have .
Sum of the zeroes =
Product of zeroes = $\left(\frac{-3}{\sqrt{2}}\right)×\left(\frac{1}{2\sqrt{2}}\right)=\frac{-3}{4}=\frac{c}{a}$

Hence, the relation is verified.

#### Question 7:

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

$2{s}^{2}-\left(1+2\sqrt{2}\right)s+\sqrt{2}$
$=2{s}^{2}-s-2\sqrt{2}s+\sqrt{2}\phantom{\rule{0ex}{0ex}}=s\left(2s-1\right)-\sqrt{2}\left(2s-1\right)\phantom{\rule{0ex}{0ex}}=\left(2s-1\right)\left(s-\sqrt{2}\right)$
To find the zeroes,
$2{s}^{2}-\left(1+2\sqrt{2}\right)s+\sqrt{2}$ = 0

$⇒$Zeroes of the given polynomial are
Now to verify the relation between the zeroes and the coefficients of the polynomial $2{s}^{2}-\left(1+2\sqrt{2}\right)s+\sqrt{2}$ we have .
Sum of the zeroes =
Product of zeroes = $\left(\frac{1}{2}\right)×\left(\sqrt{2}\right)=\frac{\sqrt{2}}{2}=\frac{c}{a}$

Hence, the relation is verified.

#### Question 8:

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

${v}^{2}+4\sqrt{3}v-15$
$={v}^{2}+5\sqrt{3}v-\sqrt{3}v-15\phantom{\rule{0ex}{0ex}}=v\left(v+5\sqrt{3}\right)-\sqrt{3}\left(v+5\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(v-\sqrt{3}\right)\left(v+5\sqrt{3}\right)$

To find the zeroes,

Zeroes of the given polynomial are .
Now to verify the relations between the zeroes and the coefficients of the polynomial ${v}^{2}+4\sqrt{3}v-15$ we have .
Sum of the zeroes = $\sqrt{3}+\left(-5\sqrt{3}\right)=-4\sqrt{3}=\frac{-b}{a}$
Product of the zeroes = $\sqrt{3}×\left(-5\sqrt{3}\right)=-15=\frac{c}{a}$

Hence, the relation is verified.

#### Question 9:

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

${y}^{2}+\frac{3}{2}\sqrt{5}y-5$

$=\frac{1}{2}\left(2{y}^{2}+3\sqrt{5}y-10\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(2{y}^{2}+4\sqrt{5}y-\sqrt{5}y-10\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2y\left(y+2\sqrt{5}\right)-\sqrt{5}\left(y+2\sqrt{5}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(y+2\sqrt{5}\right)\left(2y-\sqrt{5}\right)$
To find the zeroes:
${y}^{2}+\frac{3}{2}\sqrt{5}y-5=0$

Zeroes of given polynomial are
Now to verify the relations between the zeroes and the coefficients of the polynomial ${y}^{2}+\frac{3}{2}\sqrt{5}y-5$ we have .
Sum of zeroes = $-2\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{-3\sqrt{5}}{2}=\frac{-b}{a}$
Product of zeroes = $\left(-2\sqrt{5}\right)×\frac{\sqrt{5}}{2}=-5=\frac{c}{a}$

Hence, the relation is verified.

#### Question 10:

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

$7{y}^{2}-\frac{11}{3}y-\frac{2}{3}$
$=\frac{1}{3}\left(21{y}^{2}-11y-2\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(21{y}^{2}-14y+3y-2\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[7y\left(3y-2\right)+1\left(3y-2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(3y-2\right)\left(7y+1\right)$
To find the zeroes:
$7{y}^{2}-\frac{11}{3}y-\frac{2}{3}$ = 0

Zeroes of given polynomial are .
Now to verify the relations between the zeroes and the coefficients of the polynomial $7{y}^{2}-\frac{11}{3}y-\frac{2}{3}$ we have .
Sum of zeroes = $\frac{2}{3}+\left(\frac{-1}{7}\right)=\frac{14-3}{21}=\frac{11}{21}=\frac{-b}{a}$
Product of zeroes = $\frac{2}{3}×\left(\frac{-1}{7}\right)=\left(\frac{-2}{21}\right)=\frac{c}{a}$

Hence, the relation is verified.

#### 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)

(ii)

(iii) $-2\sqrt{3},-9$

(iv)

Quadratic polynomial when sum & product of zeroes is given is [x2 $-$ (sum of zeroes)x $+$ product of zeroes].

(i) Given: sum and product of the zeroes are . So, the polynomial is
${x}^{2}-\left(\frac{-8}{3}\right)x+\frac{4}{3}\phantom{\rule{0ex}{0ex}}={x}^{2}+\frac{8}{3}x+\frac{4}{3}$
Now,

Hence, the zeroes of the given polynomial are .

(ii) Given: sum and product of the zeroes are . So, the polynomial is
${x}^{2}-\left(\frac{21}{8}\right)x+\frac{5}{16}\phantom{\rule{0ex}{0ex}}={x}^{2}-\frac{21}{8}x+\frac{5}{16}$
Now,

Hence, the zeroes of the given polynomial are .

(iii) Given: sum and product of the zeroes are . So, the polynomial is
${x}^{2}-\left(-2\sqrt{3}\right)x+\left(-9\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+2\sqrt{3}x-9$
Now,

Hence, the zeroes of the given polynomial are .

(iv) Given: sum and product of the zeroes are . So, the polynomial is
${x}^{2}-\left(\frac{-3}{2\sqrt{5}}\right)x+\frac{\left(-1\right)}{2}\phantom{\rule{0ex}{0ex}}={x}^{2}+\frac{3}{2\sqrt{5}}x-\frac{1}{2}$

Now,

Hence, the zeroes of the given polynomial are .

#### Question 2:

Given that the zeroes of the cubic polynomial x3 – 6x2 + 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.

x3 – 6x2 + 3x + 10
Putting $x=-1$ we get,
${\left(-1\right)}^{3}-6{\left(-1\right)}^{2}+3\left(-1\right)+10\phantom{\rule{0ex}{0ex}}=-1-6-3+10\phantom{\rule{0ex}{0ex}}=-10+10\phantom{\rule{0ex}{0ex}}=0$
∵ $x=-1$ is the zero of the given polynomial.
$⇒$$\left(x+1\right)$ is the factor of the given polynomial.

$=\left(x+1\right)\left[{x}^{2}-7x+10\right]\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left({x}^{2}-5x-2x+10\right)\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left[x\left(x-5\right)-2\left(x-5\right)\right]\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left(x-2\right)\left(x-5\right)\phantom{\rule{0ex}{0ex}}$
To find the zeroes:

Zeroes of given polynomial are1, 2 and 5.
Given zeroes of the polynomial x3 – 6x2 + 3x + 10 are of the form .

Subtracting (i) from (ii) we get,
$b=3$
from (i),
$a=-4$

Hence, the values of a and b are $-4$ and 3 respectively.

#### Question 3:

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

Given $\sqrt{2}$ is a zero of the cubic polynomial $6{x}^{3}+\sqrt{2}{x}^{2}-10x-4\sqrt{2}$.
$⇒$$\left(x-\sqrt{2}\right)$ is a factor of polynomial $6{x}^{3}+\sqrt{2}{x}^{2}-10x-4\sqrt{2}$.
We have  $6{x}^{3}+\sqrt{2}{x}^{2}-10x-4\sqrt{2}$
$=\left(x-\sqrt{2}\right)\left(6{x}^{2}+7\sqrt{2}x+4\right)\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{2}\right)\left(6{x}^{2}+3\sqrt{2}x+4\sqrt{2}x+4\right)\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{2}\right)\left[3\sqrt{2}x\left(\sqrt{2}x+1\right)+4\left(\sqrt{2}x+1\right)\right]\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{2}\right)\left(3\sqrt{2}x+4\right)\left(\sqrt{2}x+1\right)$
To find the zeroes

Therefore, other two zeroes are .

#### Question 4:

Find k so that x2 $-$ 2x + k is a factor of 2x4 + x3 – 14x2 + 5x + 6. Also find all the zeroes of the two polynomials.

2x4 + x3 – 14x+ 5x + 6
$=\left(x-1\right)\left(2{x}^{3}+3{x}^{2}-11x+6\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-1\right)\left(2{x}^{2}+5x-6\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-1\right)\left(2{x}^{2}+5x-6\right)\phantom{\rule{0ex}{0ex}}=\left({x}^{2}-2x+1\right)\left(2{x}^{2}+5x-6\right)$
Factors of the given polynomial are $\left({x}^{2}-2x+1\right)\left(2{x}^{2}+5x-6\right)$.
Given x2$-$2x + k is one of the factor of the same polynomial, by comparing, $k=1$.
Hence, the value of is 1.

#### 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.

${x}^{3}-3\sqrt{5}{x}^{2}+13x-3\sqrt{5}$
$=\left(x-\sqrt{5}\right)\left({x}^{2}-2\sqrt{5}x+3\right)\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{5}\right)\left[{x}^{2}-\left(\sqrt{5}-\sqrt{2}\right)x-\left(\sqrt{5}+\sqrt{2}\right)x+\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{5}\right)\left[x\left(x-\sqrt{5}+\sqrt{2}\right)-\left(\sqrt{5}+\sqrt{2}\right)\left(x-\sqrt{5}+\sqrt{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(x-\sqrt{5}\right)\left(x-\sqrt{5}+\sqrt{2}\right)\left(x-\sqrt{5}-\sqrt{2}\right)$
To find the zeroes
${x}^{3}-3\sqrt{5}{x}^{2}+13x-3\sqrt{5}=0\phantom{\rule{0ex}{0ex}}$

Hence, the other two roots are .

#### Question 6:

For which values of a and b, are the zeroes of q(x) = x3 + 2x2 + a also the zeroes of the polynomial p(x) = x5x4 – 4x3 + 3x2 + 3x + b? Which zeroes of p(x) are not the zeroes of q(x)?

Given zeroes of q(x) are also the zeroes of p(x), then q(x) should be the factor of p(x).
So, q(x) should wholly divide p(x) with remainder 0.
p(x) = x5 – x4 – 4x3 + 3x2 + 3x + b
q
(x) = x3 + 2x2 + a
We have,
$p\left(x\right)=q\left(x\right)×\left({x}^{2}-3x+2\right)+\left[\left(-1-a\right){x}^{2}+3\left(a+1\right)x+\left(b-2a\right)\right]$
Now, for the remainder to be zero,
Coefficient of
Coefficient of $x=0⇒3\left(a+1\right)=0⇒a=-1$
And the constant term
$b-2a=0\phantom{\rule{0ex}{0ex}}b-2\left(-1\right)=0\phantom{\rule{0ex}{0ex}}b=-2$
Now, p(x) = x5 – x4 – 4x3 + 3x2 + 3x $-2$
$=\left(x-1\right)\left({x}^{4}-4{x}^{2}-x+2\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x+1\right)\left({x}^{3}-{x}^{2}-3x+2\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left({x}^{2}+x-1\right)$
and
q(x) = x3 + 2x2 $-1$
$=\left(x-1\right)\left({x}^{2}+x-1\right)$

So, are the factors of p(x) and q(x) both.
And  are the factors of p(x) only.
are the zeroes of p(x) only which are not the zeroes of q(x).

Hence, zeroes of p(x) and q(x) are same for  and zeroes of p(x), and 2 are not the zeroes of q(x).

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