Free Polynomials 02 Practice Test - 10th Grade

Question 1

$Zeroes of x^{2} - 56 x + 108 are$ ______.

both positive

both negative

one negative and one positive

both equal

SOLUTION

Solution : A

In a quadratic equation $a x^{2} + b x + c$ ,both zeroes are positive when a is positive, b is negative and c is positive.

In this case, $x^{2} - 56 x + 108$ can be factorized as $(x - 2) (x - 54)$ .

$⟹$ the roots are 2 and 54; both are positive.

Question 2

If $α, β$ and $γ$ are the zeroes of the polynomial $f (x) = a x^{3} + b x^{2} + c x + d$ , then $\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}$ is _____.

$\frac{c}{d}$

$\frac{a}{d}$

$- \frac{c}{d}$

$- \frac{b}{a}$

SOLUTION

Solution : C

Given $α, β$ and $γ$ are the zeroes of the polynomial $f (x) = a x^{3} + b x^{2} + c x + d$

$\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}$

$= \frac{β γ + α γ + α β}{(α β γ)}$ $= \frac{(\frac{c}{a})}{(- \frac{d}{a})}$ $= - \frac{c}{d}$

Question 3

When $3 x^{3} + 2 x^{2} - x + 4$ is divided by $3 + 2 x$ the remainder is $\frac{- 1}{8}$ .

True

False

SOLUTION

Solution : A

$P (x) = 3 x^{3} + 2 x^{2} - x + 4$

If $P (x)$ is divided by $(x - a)$ , then the remainder is $P (a)$

$2 x + 3 = 0 \Rightarrow x = \frac{- 3}{2}$

So, the remainder is the value of polynomial at $\frac{- 3}{2}$

Substituting $x = \frac{- 3}{2}$ in $p (x)$

$p (\frac{- 3}{2}) = 3 (\frac{- 3}{2})^{3} + 2 (\frac{- 3}{2})^{2} - (\frac{- 3}{2}) + 4$

$p (\frac{- 3}{2}) = 3 (\frac{- 27}{8}) + 2 (\frac{9}{4}) - (\frac{- 3}{2}) + 4$

$p (\frac{- 3}{2}) = \frac{- 81}{8} + \frac{9}{2} + \frac{3}{2} + 4$

$p (\frac{- 3}{2}) = \frac{- 81 + 36 + 12 + 32}{8}$

$p (\frac{- 3}{2}) = \frac{- 81 + 80}{8}$

$p (\frac{- 3}{2}) = \frac{- 1}{8}$

$∴$ The remainder is $\frac{- 1}{8} .$

Thus, the given statement is true.

Question 4

What will be the remainder when $4 x^{4} - 3 x^{3} + 2 x^{2} - x$ is divided by $x + 1$ ?

SOLUTION

Solution : D

Given $p (x) = 4 x^{4} - 3 x^{3} + 2 x^{2} - x$

To find the remainder of $\frac{4 x^{4} - 3 x^{3} + 2 x^{2} - x}{x + 1}$ ,
we will use remainder theorem.

Remainder of $\frac{p (x)}{x - a}$ is $p (a)$ .

$\Rightarrow$ Remainder of $\frac{4 x^{4} - 3 x^{3} + 2 x^{2} - x}{x + 1}$ is $p (- 1)$ .

$[∵ x + 1 = x - (- 1)]$

$p (- 1) = 4 (- 1)^{4} - 3 (- 1)^{3} + 2 (- 1)^{2} - (- 1)$

$\Rightarrow p (- 1) = 4 + 3 + 2 + 1 = 10$

Hence, the remainder of $\frac{4 x^{4} - 3 x^{3} + 2 x^{2} - x}{x + 1}$ is 10.

Question 5

Which of the following graph represents a quadratic polynomial which has sum of its zeroes is zero?

SOLUTION

Solution : D

We know that, the values at which a quadratic polynomial cuts the x-axis are its zeroes.

Option A: Polynomial cuts the x-axis at : -3, -2 ; sum of zeroes: -5

Option B: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5

Option C: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5

Option D: Polynomial cuts the x-axis at : -2, 2; sum of zeros: 0

Hence, the answer is D.

Question 6

If 1 is a zero of the polynomial p(x), then which of the following is correct?

p(1) = x

p(1) = 0

p(1) = 1

p(2) - p(1) = p(2)

SOLUTION

Solution : B and D

$If k is a zero of the polynomial p(x), then, p (k) = 0.$

$Here, 1 is a zero of the polynomial p(x).$

$\Rightarrow p (1) = 0$

$Also, p (2) - p (1) = p (2) - 0 = p (2)$

Question 7

The zeroes of $m n (x^{2} + 1) = (m^{2} + n^{2}) x$ are:

\frac{m}{n} a n d \frac{n}{m}

\frac{- m}{n} a n d \frac{n}{m}

\frac{- m}{n} a n d \frac{- n}{m}

\frac{m}{n} a n d \frac{- n}{m}

SOLUTION

Solution : A

Given polynomial, $m n (x^{2} + 1) = (m^{2} + n^{2}) x$

Let's factorise it.
$m n x^{2} - (m^{2} + n^{2}) x + m n = 0$

$\Rightarrow m n x^{2} - m^{2} x - n^{2} x + m n = 0$

$\Rightarrow m x (n x - m) - n (n x - m) = 0$

$\Rightarrow (m x - n) (n x - m) = 0$

$\Rightarrow m x - n = 0 o r n x - m = 0$

$\Rightarrow x = \frac{m}{n} o r \frac{n}{m}$

So, zeroes of the given polynomial are $\frac{m}{n}$ and $\frac{n}{m}$ .

Question 8

If $α, β$ are the zeroes of the polynomial $x^{2} - p x + 36$ and $α^{2}$ + $β^{2}$ = 9, then what is the value of p?

±6

±3

±8

±9

SOLUTION

Solution : D

Given polynomial $x^{2} - p x + 36$

On comparing with the standard form of a quadratic polynomial $a x^{2} + b x + c$ , we get
a = 1, b = -p, c = 36

Here, $α$ and $β$ are the zeroes of the polynomial.
$\Rightarrow α + β = \frac{- b}{a} = p$
and $α β = \frac{c}{a} = 36$

Now, $α^{2}$ + $β^{2} = {(α + β)}^{2}$ - 2 $α β$

$\Rightarrow$ $9 = p^{2} - 2 \times 36$ [ $∵ α^{2} + β^{2}$ = 9]
$\Rightarrow 81 = p^{2}$

$\Rightarrow p = 9 or - 9$

Question 9

Which of the following graph represents the quadratic polynomial $- x^{2} + 5 x - 6$ ?

D. None of the above

SOLUTION

Solution : C

One way of solving this question is to find the zeroes of given polynomial by conventional factorization. But we will solve it in a smart way. We just find the sum of zeroes from polynomial and compare it from the graphs, which satisfies the sum of zeroes.

From polynomial,
Sum of zeroes = $\frac{- b}{a}$ = $\frac{- 5}{- 1}$ = 5
From the graphs given , we look for graph whose sum of zeroes is 5. Thus graph having 2,3 as zeroes is the required graph.

Question 10

Graph of a quadratic polynomial is a

A. straight line

B. circle

C. parabola

D. ellipse

SOLUTION

Solution : C

Quadratic polynomial is of degree 2 and the graph of a quadratic polynomial is parabola.

Free Polynomials 02 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

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Coordinate Geometry 02
Constructions 02
Introduction to Trigonometry 02