# Free Polynomials 02 Practice Test - 9th Grade

### Question 1

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

only 1

any positive integer

any negative integer

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+x−a . 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?

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

Only one among a, b and c is zero.

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

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 is negative integer always

a is 1 always

a is 0 always

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

axn = ax0 = a

So, n has to be zero, and a should not be zero.

### Question 4

Find the factors of x3+2x2+2x+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=25a2−35a+69. The length is given as (5a−3).Find the width of the rectangle.

5a−3

5a−4

4a−5

a−4

#### SOLUTION

Solution :B

Let y be the width.

Given: Area =25a2−35a+69

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

Hence, y×(5a−3)=25a2−35a+69

y=(25a2−35a+69)÷(5a−3)

By long division method,

5a−35a−23)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯25a2−35a+69 25a2−12a − + ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ −23a+69 −23a+69 + − ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0

∴y=5a−23Width of the rectangle =5a−23

### Question 6

If a polynomial(x2+ax−c)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=2−4=−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 1012−992

#### SOLUTION

Solution :B and C

Using the algebraic Identity (a2−b2)=(a−b)(a+b)

⇒(a−b) and (a+b) are factors of a2−b2.

Consider,

(1012−992)=(101+99)(101−99)

=(200)(2)

We find that 2 and 200 both are the factors of (1012−992).

### Question 8

Factor Theorem is equivalent to Remainder Theorem when remainder is

#### 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))

1.68686868... , 0

#### 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).

True

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.