# Free Polynomials 02 Practice Test - 9th Grade

If xg+1xa is a polynomial, what could be the value of a?

A.

only 1

B.

any positive integer

C.

any negative integer

D.

any integer

#### SOLUTION

Solution : C

For an expression to be a polynomial, the exponents of x should be whole numbers. The above expression can be written as  xg+xa . For this to be a polynomial, a should be a whole number. This is possible if a is a negative integer or zero.

If ax2+bx+c is a monomial in x , what can be said about a, b and c?

A.

Atleast one of a, b and c is non - zero.

B.

Only one among a, b and c is zero.

C.

Only one among a, b and c is non - zero.

D.

All three should be non - zero.

#### SOLUTION

Solution : C

Since a monomial can contain only one term, only one of a, b and c can be non-zero to leave a single term.

If axn is a zero degree polynomial in x, what can be said about  a?

A.

a is negative integer always

B.

a is  1 always

C.

a is 0 always

D.

a is non zero

#### SOLUTION

Solution : D

A zero degree polynomial is essentially a constant. i.e., any non-zero real number.
So, if axn is a zero degree polynomial, then
axnax0 = a
So, n has to be zero, and a should not be zero.

Find the factors of x3+2x2+2x+1

A. (x+1)
B. (x2+x+1)
C. (x1)
D. (x2x+1)

#### SOLUTION

Solution : A and B

Let p(x)=(x3+2x2+2x+1) be the given polynomial
By observation, we can see that by putting
x=1, p(x)=0
x=1 is the root of p(x)
(x+1) is a factor of p(x)

To find the other factor, we perform long division of p(x) by (x+1) The two factors of p(x) are (x+1) and (x2+x+1)

Given the area of rectangle is A=25a235a+69. The length is given as (5a3).Find the width of the rectangle.

A.

5a3

B.

5a4

C.

4a5

D.

a4

#### SOLUTION

Solution : B

Let y be the width.
Given: Area =25a235a+69

We know that, area of a rectangle = Length ×breadth

Hence, y×(5a3)=25a235a+69
y=(25a235a+69)÷(5a3)
By long division method,

5a35a23)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯25a235a+69            25a212a                 +           ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯                                           23a+69                      23a+69                      +                    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯                                                  0
y=5a23Width of the rectangle =5a23

If a polynomial(x2+axc)cuts x-axis at 2 and -4, the value of a+c is

___

#### SOLUTION

Solution :

Suppose f(x) is the polynomial.
Now since it cuts x-axis at 2 and -4, the roots of f(x) are 2 and -4.
Sum of roots = a=24=2
a=2
Product of roots = c=2×4=8
c=8

Hence, (a+c)=10

Which of the following are the factors of 1012992

A. 101
B. 200
C. 2
D. 9999

#### SOLUTION

Solution : B and C

Using the algebraic Identity (a2b2)=(ab)(a+b)

(ab) and (a+b) are factors of a2b2.

Consider,
(1012992)

=(101+99)(10199)

=(200)(2)

We find that 2 and 200 both are the factors of (1012992)​.

Factor Theorem is equivalent to Remainder Theorem when remainder is

A. 2
B. 1
C. 0
D. 4

#### SOLUTION

Solution : C

If f(x) is a polynomial and is divided by (x-a), then if f(a) = k, this k will be the remainder, as stated by the Remainder Theorem. If the value of k is 0, then (x-a) is the factor of f(x), as indicated by the Factor Theorem. Hence, factor theorem is equivalent to remainder theorem when remainder is 0.

Select the values which the coefficients and exponents of a polynomial can take? (Options are given by comma separating them in the form - (coefficient,exponent))

A.

1.68686868... , 0

B. 4, -2
C. 0,0
D. -5, -5

#### SOLUTION

Solution : A and C

Any real number can be a coefficient in a polynomial. Thus, rational and irrational numbers are valid coefficients.
Exponents in polynomials can only be whole numbers.

If f(x) = (x - a)(x - b) , then if a and b are the zeros of the polynomial f(x), f(a) = f(b).

A.

True

B.

False

#### SOLUTION

Solution : A

Since a and b are the zeroes of the polynomial f(x), both f(a) = 0 and f(b) = 0, i.e. a and b will be the roots of the equation f(x) = 0.
Thus, f(a) = f(b) = 0. Hence, the given statement is true.