# Free Polynomials 03 Practice Test - 10th Grade

Find the zeros of the quadratic polynomial x28x+12.

A.

2 and 6

B.

8 and 4

C.

10 and 2

D.

4 and 5

#### SOLUTION

Solution : A

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

x28x+12=0

x26x2x+12=0

x(x6)2(x6)=0

(x6)(x2)=0

x6=0 or x2=0x=6; 2

2 and 6 are the zeros of the quadratic polynomial.

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

A.

2x2+6x+11

B.

5x2+3x+8

C.

6x2+3x+9

D.

2x2+3x+1

#### SOLUTION

Solution : D

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

P(x)=(x+2)×(2x1)+3         =x(2x1)+2(2x1)+3         =2x2x+4x2+3P(x)=2x2+3x+1

The required polynomial is 2x2+3x+1.

When we divide 3t4+5t37t2+2t+2   by   t2+3t+1, we get 3t24t+2 as quotient.

A.

True

B.

False

#### SOLUTION

Solution : A

Lets divide 3t4+5t37t2+2t+2 by t2+3t+1,

3t24t+2t2+3t+1 )¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 3t4+5t37t2+2t+2                    3t4±9t3±3t2–––––––––––––––                       4t310t2+2t4t312t24t–––––––––––––––––                                    2t2+6t+22t2±6t±2––––––––––––                                           00

Thus, the given statement is true.

Find all the common zeroes of the polynomials x3+5x29x45 and x3+8x2+15x.

A.

3

B.

5

C.

-3

D.

-5

#### SOLUTION

Solution : C and D

Let p(x)=x3+5x29x45
q(x)=x3+8x2+15x

p(x)=x3+5x29x45

p(x)=x2(x+5)9(x+5)

p(x)=(x+5)(x29)

p(x)=(x+5)(x+3)(x3)

Zeroes are -5,-3 and 3.

q(x)=x3+8x2+15x

q(x)=x(x2+8x+15)

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

Zeroes are 0, -5 and -3

Therefore, common zeroes are -5 and -3.

Find the quadratic polynomial whose sum of its zeroes (roots) is 85 and the product of the zeroes (roots)  is 75.

A.

14x2+7x+5

B.

5x2+8x+7

C.

2x28x+7

D.

5x28x+7

#### SOLUTION

Solution : B

Given that,

Sum of zeroes = 85

Product of zeroes = 75

f(x)=[(x2(sum of roots)x+(product of roots)]

Substituting the given values we get,

f(x)=[x2(8)5x+75]

f(x)=[x2+85x+75]

multiplying by 5 we get

f(x)=5x2+8x+7

Required polynomial is 5x2+8x+7.

Priya lost her homework paper on polynomials and she doesn't remember the divisor which, on dividing the polynomial x33x2+x+2 gives quotient (x2) and remainder (2x+4). Find the divisor.

A.

5x27x+9

B.

x28x+7

C.

x2x+1

D.

6x2x+11

#### SOLUTION

Solution : C

We know that,

Dividend = Divisor x Quotient +Remainder

x33x2+x+2=Divisor×(x2)+(2x+4)

Divisor×(x2)=x33x2+x+2+2x4

Divisor=(x33x2+3x2)x2

Divisor=x2x+1

If α,β and γ are the zeroes of the cubic polynomial ax3+bx2+cx+d,
then α+β+γ is equal to

A.

ca

B.

-da

C.

cd

D.

-ba

#### SOLUTION

Solution : D

If α,β and γ are the zeroes of the cubic polynomial ax3+bx2+cx+d, then

sum of its zeroes
= α+β+γ
(coefficient of x2)(coefficient of x3)=ba

Find the zeroes of the polynomial x22.

A.

2 and 5

B.

4 and 3

C.

2 and 2

D.

5 and 2

#### SOLUTION

Solution : C

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

x22=0

Using  the identity, a2b2=(a+b)(ab)

(x+2)(x2)=0

x+2=0 and x2=0

x=2 and x=2

Hence, the zeroes are 2,2

Find the quadratic polynomial whose zeroes are 6+33 and 633 .

A.

x216x+9

B.

3x212x+11

C.

6x24x+1

D.

7x24x+11

#### SOLUTION

Solution : B

Let α & β be the roots.
Sum of zeroes,  α+β=6+33+633                                       =123=4

Product of zeroes, αβ=6+33×633=62(3)29αβ=3639=339=113

Required polynomial f(x)=[(x2(α+β)x+(αβ)]        =[(x24x+113]

Multiplying by 3

f(x)=3x212x+11

the required polynomial is 3x212x+11.

The product of zeros of cubic polynomial x33x2x+3 is

A. -3
B. -1
C. 3
D. 1

#### SOLUTION

Solution : A

For a cubic polynomial ax3+bx2+cx+d
Sum of zeros = ba
Product of zeros = da
Product of zeros is 31 = -3