Free Polynomials 03 Practice Test - 10th Grade

Question 1

$Find the zeros of the quadratic polynomial x^{2} - 8 x + 12.$

$2 and 6$

$8 and 4$

$10 and 2$

$4 and 5$

SOLUTION

Solution : A

To find the zeroes, equate the polynomial to zero and then factorise.

$∴ x^{2} - 8 x + 12 = 0$

$\Rightarrow x^{2} - 6 x - 2 x + 12 = 0$

$\Rightarrow x (x - 6) - 2 (x - 6) = 0$

$\Rightarrow (x - 6) (x - 2) = 0$

$\Rightarrow x - 6 = 0 or x - 2 = 0 \Rightarrow x = 6; 2$

$2 and 6 are the zeros of the quadratic polynomial.$

Question 2

When a polynomial is divided by (x+2), the quotient and remainder are (2x-1) and 3 respectively. Find the polynomial.

$2 x^{2} + 6 x + 11$

$5 x^{2} + 3 x + 8$

$6 x^{2} + 3 x + 9$

$2 x^{2} + 3 x + 1$

SOLUTION

Solution : D

We know that, by divison algorithm,
Dividend = Divisor x Quotient + Remainder.

$∴ P (x) = (x + 2) \times (2 x - 1) + 3 = x (2 x - 1) + 2 (2 x - 1) + 3 = 2 x^{2} - x + 4 x - 2 + 3 P (x) = 2 x^{2} + 3 x + 1$

$∴ The required polynomial is 2 x^{2} + 3 x + 1.$

Question 3

When we divide $3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2$ by $t^{2} + 3 t + 1$ , we get $3 t^{2} - 4 t + 2$ as quotient.

True

False

SOLUTION

Solution : A

Lets divide $3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2 by t^{2} + 3 t + 1$ ,

$3 t^{2} - 4 t + 2 t^{2} + 3 t + 1) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 3 t^{4} + 5 t^{3} - 7 t^{2} + 2 t + 2_{-} 3 t^{4} \pm 9 t^{3} \pm 3 t^{2} - --------------- - - 4 t^{3} - 10 t^{2} + 2 t \mp 4 t^{3} \mp 12 t^{2} \mp 4 t - ----------------- - 2 t^{2} + 6 t + 2_{-} 2 t^{2} \pm 6 t \pm 2 - ------------ - 00$

Thus, the given statement is true.

Question 4

Find all the common zeroes of the polynomials $x^{3} + 5 x^{2} - 9 x - 45$ and $x^{3} + 8 x^{2} + 15 x$ .

-3

-5

SOLUTION

Solution : C and D

Let $p (x) = x^{3} + 5 x^{2} - 9 x - 45$
$q (x) = x^{3} + 8 x^{2} + 15 x$

$p (x) = x^{3} + 5 x^{2} - 9 x - 45$

$p (x) = x^{2} (x + 5) - 9 (x + 5)$

$p (x) = (x + 5) (x^{2} - 9)$

$p (x) = (x + 5) (x + 3) (x - 3)$

$\Rightarrow$ Zeroes are -5,-3 and 3.

$q (x) = x^{3} + 8 x^{2} + 15 x$

$q (x) = x (x^{2} + 8 x + 15)$

$q (x) = x (x + 5) (x + 3)$

$\Rightarrow$ Zeroes are 0, -5 and -3

Therefore, common zeroes are -5 and -3.

Question 5

Find the quadratic polynomial whose sum of its zeroes (roots) is $- \frac{8}{5}$ and the product of the zeroes (roots) is $\frac{7}{5}$ .

$14 x^{2} + 7 x + 5$

$5 x^{2} + 8 x + 7$

$2 x^{2} - 8 x + 7$

$5 x^{2} - 8 x + 7$

SOLUTION

Solution : B

Given that,

Sum of zeroes = $\frac{- 8}{5}$

Product of zeroes = $\frac{7}{5}$

Required quadratic polynomial is,

$f (x) = [(x^{2} - (s u m o f r o o t s) x + (p r o d u c t o f r o o t s)]$

Substituting the given values we get,

$f (x) = [x^{2} - \frac{(- 8)}{5} x + \frac{7}{5}]$

$f (x) = [x^{2} + \frac{8}{5} x + \frac{7}{5}]$

multiplying by 5 we get

$f (x) = 5 x^{2} + 8 x + 7$

$∴$ Required polynomial is $5 x^{2} + 8 x + 7$ .

Question 6

Priya lost her homework paper on polynomials and she doesn't remember the divisor which, on dividing the polynomial $x^{3} - 3 x^{2} + x + 2$ gives quotient $(x - 2)$ and remainder $(- 2 x + 4) .$ Find the divisor.

$5 x^{2} - 7 x + 9$

$x^{2} - 8 x + 7$

$x^{2} - x + 1$

$6 x^{2} - x + 11$

SOLUTION

Solution : C

We know that,

Dividend = Divisor x Quotient +Remainder

$x^{3} - 3 x^{2} + x + 2 = D i v i s o r \times (x - 2) + (- 2 x + 4)$

$\Rightarrow D i v i s o r \times (x - 2) = x^{3} - 3 x^{2} + x + 2 + 2 x - 4$

$\Rightarrow D i v i s o r = \frac{(x^{3} - 3 x^{2} + 3 x - 2)}{x - 2}$

$∴ D i v i s o r = x^{2} - x + 1$

Question 7

If $α, β$ and $γ$ are the zeroes of the cubic polynomial $a x^{3} + b x^{2} + c x + d$ ,
then $α + β + γ$ is equal to

$\frac{c}{a}$

- $\frac{d}{a}$

$\frac{c}{d}$

- $\frac{b}{a}$

SOLUTION

Solution : D

If $α, β$ and $γ$ are the zeroes of the cubic polynomial $a x^{3} + b x^{2} + c x + d$ , then

sum of its zeroes
= $α + β + γ$
= $- \frac{(c o e f f i c i e n t o f x^{2})}{(c o e f f i c i e n t o f x^{3})} = - \frac{b}{a}$

Question 8

$Find the zeroes of the polynomial x^{2} - 2.$

$\sqrt{2} and - \sqrt{5}$

$4 and 3$

$\sqrt{2} and - \sqrt{2}$

$- 5 and 2$

SOLUTION

Solution : C

$To find the zeroes, equate the given polynomial to zero.$

$x^{2} - 2 = 0$

$Using the identity, a^{2} - b^{2} = (a + b) (a - b)$

$\Rightarrow (x + \sqrt{2}) (x - \sqrt{2}) = 0$

$\Rightarrow x + \sqrt{2} = 0 and x - \sqrt{2} = 0$

$\Rightarrow x = - \sqrt{2} and x = \sqrt{2}$

$Hence, the zeroes are \sqrt{2}, - \sqrt{2}$ .

Question 9

Find the quadratic polynomial whose zeroes are $\frac{6 + \sqrt{3}}{3} and \frac{6 - \sqrt{3}}{3} .$

$x^{2} - 16 x + 9$

$3 x^{2} - 12 x + 11$

$6 x^{2} - 4 x + 1$

$7 x^{2} - 4 x + 11$

SOLUTION

Solution : B

Let $α & β$ be the roots.
Sum of zeroes, $α + β = \frac{6 + \sqrt{3}}{3} + \frac{6 - \sqrt{3}}{3}$ $= \frac{12}{3} = 4$

Product of zeroes, $α β = \frac{6 + \sqrt{3}}{3} \times \frac{6 - \sqrt{3}}{3} = \frac{6^{2} - (\sqrt{3})^{2}}{9} α β = \frac{36 - 3}{9} = \frac{33}{9} = \frac{11}{3}$

Required polynomial $f (x) = [(x^{2} - (α + β) x + (α β)] = [(x^{2} - 4 x + \frac{11}{3}]$

Multiplying by 3

$\Rightarrow f (x) = 3 x^{2} - 12 x + 11$

$∴$ the required polynomial is $3 x^{2} - 12 x + 11$ .

Question 10

The product of zeros of cubic polynomial $x^{3} - 3 x^{2} - x + 3$ is

A. -3

B. -1

C. 3

D. 1

SOLUTION

Solution : A

For a cubic polynomial $a x^{3} + b x^{2} + c x + d$
Sum of zeros = $\frac{- b}{a}$
Product of zeros = $\frac{- d}{a}$
$∴$ Product of zeros is $\frac{- 3}{1}$ = -3

Free Polynomials 03 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Related ExamsTriangles 03Introduction to Trigonometry 03Surface Areas and Volumes 03

Related Exams
Triangles 03
Introduction to Trigonometry 03
Surface Areas and Volumes 03