Free Polynomials 01 Practice Test - 9th Grade
Question 1
Degree of a constant polynomial is _____.
1
0
Any natural number
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.
Question 2
If x+x−1=11, evaluate x2+x−2.
SOLUTION
Solution : A
x+x−1=11
(x+x−1)2=112 [Squaring both the sides]
⇒(x)2+(x−1)2+2(x)(x−1)=121
⇒ x2+x−2+2=121 [∵(x)(x−1)=x0=1]
⇒ x2+x−2=121−2=119
Question 3
The polynomial ax3+bx2+x–6 has (x+2) as a factor and leaves a remainder 4 when divided by (x–2). Find a and b.
SOLUTION
Solution : A
Let
p(x)=ax3+bx2+x−6By 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+4b−8=0
∴−2a+b=2...(i)Also when p(x) is divided by (x−2) the remainder is 4.
∴p(2)=4
⇒a(2)3+b(2)2+2−6=4
⇒8a+4b+2−6=4
⇒8a+4b=8
⇒2a+b=2...(ii)Adding equations (i) and (ii), we get
(−2a+b)+(2a+b)=2+2
⇒2b=4⇒b=2Putting b=2 in (i) we get
−2a+2=2
⇒−2a=0⇒a=0
Hence,
a=0 and b=2
Question 4
Find the roots of the equation: x-5=0
5
-5
0
Not defined.
SOLUTION
Solution : A
x−5=0 signifies that (x−5) 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.
Question 5
If a+b+c=8 and ab+bc+ca=20, find the value of a3+b3+c3–3abc.
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+c3−3abc=(a+b+c)(a2+b2+c2−(ab+bc+ca))
a3+b3+c3−3abc=8×(24−20)=4×8=32
Thus, a3+b3+c3−3abc=32
Question 6
The polynomials ax3+4x2+3x–4 and x3–4x+a leave the same remainder when divided by (x–3). Find the value of a.
SOLUTION
Solution : A
Given polynomials:
p(x)=ax3+4x2+3x−4
q(x)=x3−4x+a
Using remainder theorm, the remainders when p(x) and q(x) are divided by (x−3) are p(3) and q(3) respectively.
p(3)=a×(3)3+4×(3)2+3×3−4
p(3)=27a+41
q(3)=(3)3−4×3+a
q(3)=15+a
Given that the remainder is the same.
⇒p(3)=q(3)
27a+41=15+a
26a=26
a=1
Question 7
If (x13+y13+z13=0), then -
x+y+z=0
(x+y+z)3=27xyz
x+y+z=3x13y13z13
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
Question 8
If (x4+x2y+y2) is one of the factors of an expression which is the difference of two cubes, then the other factor is
SOLUTION
Solution : A
Let two numbers be a and b.
The difference of their cubes is a3–b3 and hence this is the expression we need.
This expression can be factorised as a3–b3=(a−b)(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 a–b.
Hence, other factor is x2−y.
Question 9
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.
Question 10
If x100+2x99+k is divisible by (x+1), then the value of k is
1
-3
3
-2
SOLUTION
Solution : A
Since (x+1) is a factor, x=−1 will make the given expression zero.
∴(−1)100+2×(−1)99+k=0⇒1−2+k=0⇒k=1