# Free Polynomials 02 Practice Test - 10th Grade

Zeroes of x256x+108 are______.

A.

both positive

B.

both negative

C.

one negative and one positive

D.

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, x256x+108 can be factorized as (x2)(x54).

the roots are 2 and  54; both are positive.

If α,β and γ are the zeroes of the polynomial f(x)=ax3+bx2+cx+d, then 1α+1β+1γ is _____.

A.

cd

B.

C.

cd

D.

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

When 3x3+2x2x+4 is divided by 3+2x the remainder is 18.

A.

True

B.

False

#### SOLUTION

Solution : A

P(x)=3x3+2x2x+4

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

2x+3=0x=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.

What will be the remainder when 4x43x3+2x2x is divided by x+1?

A.

1

B.

0

C.

7

D.

10

#### SOLUTION

Solution : D

Given p(x)=4x43x3+2x2x

To find the remainder of  4x43x3+2x2xx+1,
we will use remainder theorem.

Remainder of p(x)xa is p(a).

Remainder of  4x43x3+2x2xx+1 is p(1).

[x+1=x(1)]

p(1)=4(1)43(1)3+2(1)2(1)

p(1)=4+3+2+1=10

Hence, the remainder of 4x43x3+2x2xx+1 is 10.

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

A.

A) B.

B) C.

C) D.

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

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

A.

p(1) = x

B.

p(1) = 0

C.

p(1) = 1

D.

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)

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

A. mn and nm
B. mn and nm
C. mn and nm
D. mn and nm

#### SOLUTION

Solution : A

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

Let's factorise it.
mnx2(m2+n2)x+mn=0

mnx2m2xn2x+mn=0

mx(nxm)n(nxm)=0

(mxn)(nxm)=0

mxn=0 or nxm=0

x=mn or nm

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

If α,β are the zeroes of the polynomial x2px+36 and α2+β2 = 9, then what is the value of p?

A.

±6

B.

±3

C.

±8

D.

±9

#### SOLUTION

Solution : D

Given polynomial x2px+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=p22×36  [α2+β2 = 9]
81=p2

p=9  or 9

Which of the following graph represents the quadratic polynomial x2+5x6?

A. B. C. 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 = ba = 51 = 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.

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.