# Free Polynomials 03 Practice Test - 9th Grade

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

A.

1, 2, 3,...

B.

1, 3, 5...

C.

2, 4, 6...

D.

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.

If x+1x=2, find the value of x64+1x64

A.

32

B.

16

C.

8

D.

2

#### SOLUTION

Solution : D

x+1x=2

Squaring both sides, we get x2+1x2+2.x.1x=4
So, x2+1x2=42=2
Again Squaring both sides, we get x4+1x4+2.x2.1x2=4
x4+1x4=42=2

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

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

A.

3xyz

B.

3xyz

C.

2xyz

D.

2xyz

#### SOLUTION

Solution : A

We know that

x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx)(1)

According to question, (x2+y2+z2)=(xy+yz+zx).
RHS will be zero in equation (1)

So,x3+y3+z3=3xyz

If (x1) is a factor of ax3bx2+cx, what is the value of a3b3+c3?

A.

2abc

B.

-2abc

C.

3abc

D.

-3abc

#### SOLUTION

Solution : D

p(x)=ax3bx2+cx

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

p(1) = 0

p(1)=ab+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

If (x2y2)=18 and (xy)=3, find the value of 16x2y2.

A. 27
B. 81
C. 243
D. 729

#### SOLUTION

Solution : D

(x2y2)=(xy)(x+y)=18

(xy)=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

In a2x2+ax, if 'a' is constant and 'x' is a variable, this expression would become a polynomial in ___ variable(s). (Write in words)

#### 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.

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+xa
Since g = a,
xg+1xa=xa+xa
For the expression  xa+xa 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+xa

xg+1xa=x0+x0

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.

Factorize : x27x+12

A.

(x-3)(x-4)

B. (x+3)(x-4)
C. (x-3)(x+4)
D. (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 .
x27x+12
=x24x3x+12
=x(x4)3(x4)
=(x3)(x4)

Thus x27x+12 can be factorised as (x3)(x4)

p(x)=x3+8x27x+12 and g(x)=x1. 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), (ab)=?.

A.

1

B. 2
C. 3
D. 4

#### SOLUTION

Solution : B

Remainder =r(x)=p(1)=13+8×127×1+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+8x27x+12
g(x)=(x1)
Performing long division

x2+9x+2x1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x3+8x27x+12       x3x2       ()     (+)           ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯                                       9x27x                  9x29x              ()      (+)               ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯                                            +2x+12                         2x2                     ()    (+)                      ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯                                                    14

q(x)=p(x)g(x)=x2+9x+2
So, degree of q(x)=a=2
Hence, ab=20=2

The value of (2x3y)3+(x+3y)3+3(2x3y)2(x+3y)+3(2x3y)(x+3y)2=

A. 27x3
B. 29x3+(x+3y)3
C. (2x3y)3+(x+3y)3
D. 0

#### SOLUTION

Solution : A

Substitute 2x3y=a and (x+3y)=b
(2x3y)3+(x+3y)3+3(2x3y)2(x+3y)+3(2x3y)(x+3y)2
=a3+b3+3a2b+3ab2
=(a+b)3
=(2x3y+x+3y)3
=(3x)3
=27x3