Free Polynomials 02 Practice Test - 9th Grade 

Question 1

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.

Question 2

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.

Question 3

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.

Question 4

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)

Question 5

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

Question 6

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

Question 7

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)​.

Question 8

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.

Question 9

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.

Question 10

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.