# Free Polynomials 03 Practice Test - 9th Grade

### Question 1

Which of the following given sequences can be represented as a polynomial in one variable?

1, 2, 3,...

1, 3, 5...

2, 4, 6...

1, 4, 9...

#### SOLUTION

Solution :A, B, C, and D

1, 2, 3... sequence can be represented using the polynomial x.

1, 3, 5... sequence can be represented using the polynomial 2x+1.

2, 4, 6... sequence can be represented using the polynomial 2x.

1, 4, 9... sequence can be represented using the polynomial x2.

All of these are polynomials in one variable i.e., x.

### Question 2

If x+1x=2, find the value of x64+1x64

32

16

8

2

#### SOLUTION

Solution :D

x+1x=2

Squaring both sides, we get x2+1x2+2.x.1x=4

So, x2+1x2=4−2=2

Again Squaring both sides, we get x4+1x4+2.x2.1x2=4

⇒x4+1x4=4−2=2

Similarly, x8+1x8=x32+1x32=x64+1x64=2

### Question 3

If (x2+y2+z2)=(xy+yz+zx), what is the value of x3+y3+z3?

3xyz

−3xyz

2xyz

−2xyz

#### SOLUTION

Solution :A

We know that

x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)−−−(1)

According to question, (x2+y2+z2)=(xy+yz+zx).

∴ RHS will be zero in equation (1)

So,x3+y3+z3=3xyz

### Question 4

If (x−1) is a factor of ax3−bx2+cx, what is the value of a3−b3+c3?

2abc

-2abc

3abc

-3abc

#### SOLUTION

Solution :D

p(x)=ax3−bx2+cx

If (x - 1) is a factor then x = 1 is the zero of polynomial

p(1) = 0p(1)=a−b+c=0

It can be written as a+(−b)+c=0

We know that if a+b+c=0, then a3+b3+c3=3abc

So, a3+(−b)3+c3=3a(−b)c=−3abc

### Question 5

If (x2−y2)=18 and (x−y)=3, find the value of 16x2y2.

#### SOLUTION

Solution :D

(x2−y2)=(x−y)(x+y)=18

(x−y)=3−−−−(i)

(x+y)=183=6−−−−(ii)Adding equation (i) and (ii) we get

2x=9

x=92

Substituting the value of x in equation (ii) we get

y=32

16x2y2=16×(92)2×(32)2=729

### Question 6

In a2x2+ax, if 'a' is constant and 'x' is a variable, this expression would become a polynomial in

#### SOLUTION

Solution :In this expression, a2x2 and ax are the two terms. Since it is given that 'a' is a constant and 'x' is a variable, we can see that there is only one variable 'x' in the polynomial thus making it a polynomial in one variable.

### Question 7

If xg+1xa is a polynomial in one variable and if it is given that g = a, for what value of 'a' is this a polynomial?

#### SOLUTION

Solution :xg+1xa=xg+x−a

Since g = a,

xg+1xa=xa+x−a

For the expression xa+x−a to be a polynomial, a and -a must be whole number so that the expression 1xa = 1 , hence a should be equal to 0 .

Thus, a = 0.

When a=0, the given polynomial will get simplified further as:

xg+1xa=xa+x−a

⇒ xg+1xa=x0+x−0

⇒ xg+1xa=x0+x0

⇒ xg+1xa=1+1=2

'2' is a constant polynomial in one variable with degree zero . Because 2=2x0.

So, 'a' must be equal to zero so that given expression is a polynomial in variable.

### Question 8

Factorize : x2−7x+12

(x-3)(x-4)

#### SOLUTION

Solution :A

We have to break up 7x in the given expression into two parts in such a way that their sum is 7x and the product of their coefficients is 12 .

x2−7x+12

=x2−4x−3x+12

=x(x−4)−3(x−4)

=(x−3)(x−4)

Thus x2−7x+12 can be factorised as (x−3)(x−4)

### Question 9

p(x)=x3+8x2−7x+12 and g(x)=x−1. If p(x) is divided by g(x), it gives q(x) and r(x) as quotient and remainder respectively. If a is the degree of q(x) and b is the degree of r(x), (a−b)=?.

1

#### SOLUTION

Solution :B

Remainder =r(x)=p(1)=13+8×12−7×1+12=14

The degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials.

So, Degree of r(x)=b=0

q(x)=p(x)g(x)

Now , p(x)=x3+8x2−7x+12

g(x)=(x−1)

Performing long division

x2+9x+2x−1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x3+8x2−7x+12 x3−x2 (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 9x2−7x 9x2−9x (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ +2x+12 2x−2 (−) (+) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 14

∴q(x)=p(x)g(x)=x2+9x+2

So, degree of q(x)=a=2

Hence, a−b=2−0=2

### Question 10

The value of (2x−3y)3+(x+3y)3+3(2x−3y)2(x+3y)+3(2x−3y)(x+3y)2=

#### SOLUTION

Solution :A

Substitute 2x−3y=a and (x+3y)=b

(2x−3y)3+(x+3y)3+3(2x−3y)2(x+3y)+3(2x−3y)(x+3y)2

=a3+b3+3a2b+3ab2

=(a+b)3

=(2x−3y+x+3y)3

=(3x)3

=27x3