# Free Polynomials 02 Practice Test - 10th Grade

### Question 1

Zeroes of x2−56x+108 are______.

both positive

both negative

one negative and one positive

both equal

#### SOLUTION

Solution :A

In a quadratic equation ax2+bx+c,both zeroes are positive when a is positive, b is negative and c is positive.

In this case, x2−56x+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)=ax3+bx2+cx+d, then 1α+1β+1γ is _____.

cd

ad

−cd

−ba

#### SOLUTION

Solution :C

Given α,β and γ are the zeroes of the polynomial f(x)=ax3+bx2+cx+d

1α+1β+1γ

=βγ+αγ+αβ(αβγ) =(ca)(−da) =−cd

### Question 3

When 3x3+2x2−x+4 is divided by 3+2x the remainder is −18.

True

False

#### SOLUTION

Solution :A

P(x)=3x3+2x2−x+4

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

2x+3=0⇒x=−32

So, the remainder is the value of polynomial at −32

Substituting x=−32 in p(x)

p(−32)=3(−32)3+2(−32)2−(−32)+4

p(−32)=3(−278)+2(94)−(−32)+4

p(−32)=−818+92+32+4

p(−32)=−81+36+12+328

p(−32)=−81+808

p(−32)=−18

∴ The remainder is −18.

Thus, the given statement is true.

### Question 4

What will be the remainder when 4x4−3x3+2x2−x is divided by x+1?

1

0

7

10

#### SOLUTION

Solution :D

Given p(x)=4x4−3x3+2x2−x

To find the remainder of 4x4−3x3+2x2−xx+1,

we will use remainder theorem.

Remainder of p(x)x−a is p(a).

⇒ Remainder of 4x4−3x3+2x2−xx+1 is p(−1).[∵x+1=x−(−1)]

p(−1)=4(−1)4−3(−1)3+2(−1)2−(−1)

⇒p(−1)=4+3+2+1=10Hence, the remainder of 4x4−3x3+2x2−xx+1 is 10.

### Question 5

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

A)

B)

C)

D)

#### 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).

⇒p(1)=0

Also, p(2)−p(1)=p(2)−0=p(2)

### Question 7

The zeroes of mn(x2+1)=(m2+n2)x are:

#### SOLUTION

Solution :A

Given polynomial, mn(x2+1)=(m2+n2)x

Let's factorise it.

mnx2−(m2+n2)x+mn=0

⇒mnx2−m2x−n2x+mn=0

⇒mx(nx−m)−n(nx−m)=0

⇒(mx−n)(nx−m)=0

⇒mx−n=0 or nx−m=0

⇒x=mn or nm

So, zeroes of the given polynomial are mn and nm.

### Question 8

If α,β are the zeroes of the polynomial x2−px+36 and α2+β2 = 9, then what is the value of p?

±6

±3

±8

±9

#### SOLUTION

Solution :D

Given polynomial x2−px+36

On comparing with the standard form of a quadratic polynomial ax2+bx+c, we get

a = 1, b = -p, c = 36

Here, α and β are the zeroes of the polynomial.

⇒α+β=−ba=p

and αβ=ca=36

Now, α2+β2=(α+β)2 - 2αβ

⇒ 9=p2−2×36 [∵α2+β2 = 9]

⇒81=p2

⇒p=9 or −9

### Question 9

Which of the following graph represents the quadratic polynomial −x2+5x−6?

#### 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 = −ba = −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

#### SOLUTION

Solution :C

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