# Free Polynomials 01 Practice Test - 9th Grade

Degree of a constant polynomial is _____.

A.

1

B.

0

C.

Any natural number

D.

Not defined

#### SOLUTION

Solution : B

As we know that any variable raised to the power of zero yields 1, a constant polynomial can be represented as ax0, where, a is the constant and x is the variable of the polynomial. Thus the degree of a constant polynomial is 0.

If x+x1=11, evaluate x2+x2.

A. 119
B. 120
C. 121
D. 122

#### SOLUTION

Solution : A

x+x1=11

(x+x1)2=112  [Squaring both the sides]

(x)2+(x1)2+2(x)(x1)=121

x2+x2+2=121   [(x)(x1)=x0=1]

x2+x2=1212=119

The polynomial ax3+bx2+x6 has (x+2) as a factor and leaves a remainder 4 when divided by (x2). Find a and b.

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

#### SOLUTION

Solution : A

Let
p(x)=ax3+bx2+x6

By using factor theorem, (x+2) can be a factor of p(x) only when p(2)=0
p(2)=a(2)3+b(2)2+(2)6=0
8a+4b8=0
2a+b=2...(i)

Also when p(x) is divided by (x2) the remainder is 4.

p(2)=4
a(2)3+b(2)2+26=4
8a+4b+26=4
8a+4b=8
2a+b=2...(ii)

Adding equations (i) and (ii), we get
(2a+b)+(2a+b)=2+2
2b=4b=2

Putting b=2 in (i) we get

2a+2=2
2a=0a=0
Hence,
a=0 and b=2

Find the roots of the equation: x-5=0

A.

5

B.

-5

C.

0

D.

Not defined.

#### SOLUTION

Solution : A

x5=0 signifies that (x5) is a polynomial whose value is zero for a certain value of x and this x will be the root of the given equation

at x=5 the value of expression x-5 is 0.
Hence, the root of the given equation is 5.

If a+b+c=8 and ab+bc+ca=20, find the value of a3+b3+c33abc.

A. 16
B. 32
C. 48
D. 64

#### SOLUTION

Solution : B

Since (a+b+c)2=a2+b2+c2+2(ab+bc+ca)
a+b+c=8 and ab+bc+ca=20
(8)2=a2+b2+c2+2×20
64=a2+b2+c2+40
a2+b2+c2=24
We now use the following identity:
a3+b3+c33abc=(a+b+c)(a2+b2+c2(ab+bc+ca))
a3+b3+c33abc=8×(2420)=4×8=32
Thus, a3+b3+c33abc=32

The polynomials ax3+4x2+3x4 and x34x+a leave the same remainder when divided by (x3). Find the value of a.

A. 1
B. 2
C. 3
D. 4

#### SOLUTION

Solution : A

Given polynomials:
p(x)=ax3+4x2+3x4
q(x)=x34x+a

Using remainder theorm,​​​ t​he remainders when p(x) and q(x) are divided by (x3) are p(3) and q(3) respectively.
p(3)=a×(3)3+4×(3)2+3×34
p(3)=27a+41

q(3)=(3)34×3+a
q(3)=15+a

Given that the remainder is the same.
p(3)=q(3)
27a+41=15+a
26a=26
a=1

If (x13+y13+z13=0), then -

A.

x+y+z=0

B.

(x+y+z)3=27xyz

C.

x+y+z=3x13y13z13

D.

x3+y3+z3=27xyz

#### SOLUTION

Solution : B and C

Let x13=a,y13=b,z13=c
It is given that a+b+c=0
a+b=c
Cubing both sides, we get,
(a+b)3=(c)3
a3+b3+3ab(a+b)=(c)3
a3+b3+3ab(c)=c3
a3+b3+c3=3abc
Putting back the values of a,b,c in the above equation, we have

(x13)3+(y13)3+(z13)3=3x13y13z13
x+y+z=3x13y13z13
(x+y+z)3=(3x13y13z13)3
=27xyz

If (x4+x2y+y2) is one of the factors of an expression which is the difference of two cubes, then the other factor is

A. x2y
B. x2y2
C. xy2
D. xy

#### SOLUTION

Solution : A

Let two numbers be a and b.
The difference of their cubes is a3b3 and hence this is the expression we need.

This expression can be factorised as a3b3=(ab)(a2+ab+b2).

Now, consider the given factor, x4+x2y+y2=(x2)2+x2y+(y)2

This is in the form of a2+ab+b2 where a=x2,b=y, which is similar to the factor given.

Therefore, the other factor is  of the form ab.

Hence, other factor is x2y.

If a polynomial cuts the x-axis at 3 points and y-axis at 2 points, then the number of zeroes of the polynomial would be

___

#### SOLUTION

Solution :

The point where the polynomial cuts the x-axis is the zero of the polynomial. Since the given polynomial cuts the x-axis at 3 points, the number of zeroes is 3.

If x100+2x99+k is divisible by (x+1), then the value of k is

A.

1

B.

-3

C.

3

D.

-2

#### SOLUTION

Solution : A

Since (x+1) is a factor, x=1 will make the given expression zero.

(1)100+2×(1)99+k=012+k=0k=1