MCQ for Class 9 Mathematics Polynomials
Polynomials MCQ Questions Class 9 Mathematics with Answers
Question. In 2 + x + x2 the coefficient of x2 is: Ans is 1
(a) 2
(b) 1
(c) – 2
(d) –1
Question. In 2 – x2 + x3 the coefficient of x2 is: Ans is -1
(a) 2
(b) 1
(c) – 2
(d) –1
Question. In πx2/2 +x+10 the coefficient of x2 is: Ans is π/2
(a) π/2
(b) 1
(c) – Ï€/2
(d) –1
Question. The degree of 5t – 7 is:
(a) 0
(b) 1
(c) 2
(d) 3
Ans. Clearly, the given the given expression 5t−7 is a linear polynomial, means the highest power of t is 1. Therefore the degree of the given polynomial is 1.
Question. The degree of 4 – y2 is:
(a) 0
(b) 1
(c) 2
(d) 3
Ans. The degree of a polynomial is the highest degree of its terms. Here y2 has degree of 2.
Hence, the degree of the polynomial 4−y2 is 2
Question. The degree
of 3 is:
Ans. 3 can be written as 3x0 is a
constant polynomial.
i.e. the highest power of x is 0. Therefore, the degree of the polynomial 3 is 0
(a) 0
(b) 1
(c) 2
(d) 3
Question. The value of p(x) = 5x – 4x2 + 3 for x = 0 is: Ans. = 5(0) - 4(0)2 + 3 = 0 - 0 + 3 = 3
(a) 3
(b) 2
(c) – 3
(d) – 2
Question. The value of p(x) = 5x – 4x2 + 3 for x = – 1 is: Ans. 5(-1) - 4(-1)² + 3 = -5 - 4 + 3 = -6
(a) 6
(b) –6
(c) 3
(d) – 3
Question. The value of p(x) = (x – 1)(x + 1) for p(1) is:
Ans. Put the value p(1) in the given equation p(1)=(1-1)(1+1)=0×2=0∴p(1)=0
(a) 1
(b) 0
(c) 2
(d) – 2
Question. The value of p(t) = 2 + t + 2t2 – t3 for p(0) is:
Ans. Put the value of p(0) in the given equation p(0)=2+(0)+2x02-03=2 ∴p(0)=2
(a) 1
(b) 2
(c) – 1
(d) 3
Question. The value of p(t) = 2 + t + 2t2 – t3 for p(2) is
Ans. Put the value of p(2) in the given equation =2+2+2x(2)2-23=2+2+8-8=4
(a) 4
(b) –4
(c) 6
(d) 7
Question. The value of p(y) = y2 – y +1 for p(0) is: Ans. p(0)=02−0+1 = 1
(a) –1
(b) 3
(c) –2
(d) 1
Q. The zeroes of p(x) = 2x – 7 is
Let p(x) = 0
2x - 7 = 0
2x = 7
x = 7/2
Q. the zero of p(x) = 9x + 4 is Let p(x) = 0
9x + 4 = 0
X = -4/9
Q. Which are the zeroes of p(x) = x2-1
Ans. x2−1=0 ⇒x2−12=0 ⇒(x−1)(x+1)=0 (∵a2−b2=(a−b)(a+b)) ⇒(x−1)=0,(x+1)=0 ⇒x=1,x=−1 Hence, the zeroes of the polynomial x2−1 are 1,−1.
Q. which are the zeros of p(x)=(x-1)(x-2)
Ans. Given that p(x)=(x-1)(x-2)
zero’s of polynomial p(x) = 0 i.e. (x-1)(x-2)
= 0
x-1 = 0, x- 2 = 0
x = 0+1, x = 0+2
x = 1 and 2 so option c is correct
Q. which are the zeros of p(x)=(x-1)(x-2)(x-3)(x-4)
Given
that p(x)=(x-1)(x-2)(x-3)(x-4)
if any one term in p(x) is zero
then value of p(x)=0 so if x=1 or x=2 or x=3
or x=4
p(x)=0 so the zeros are
1,2,3,4
Q. which one of the following is the zero of p(x) = lx + m
Ans. P(x) = lx + m
Zero’s of polynomial p(x) i.e. p(x) = 0 or lx + m = 0
lx = - m
x = -m/l option c is correct
Q. which one of the following is the zero of p(x)=5x-Ï€
Ans. P(x) = 5x – Ï€
Zero’s of p(x), we get p(x) = 0
i.e. 5x – Ï€ = 0
5x = π or x = π/5 option (d) is correct
Q. On dividing x3 + 3x2 + 3x + 1 by x we get remainder
Ans. P(x) = x3 + 3x2 + 3x + 1
Apply remainder theorem we get X = 0
Replacing x by 0 we get
P(0) = (o)3 + 3(0)2 + 3(0) + 1
= 0 + 0 + 0 + 1
P(0) = 1 so option (a) is the answer
Q. On dividing x3+3x2+3x+1 by x+Ï€
Let p(x) = x3+3x2+3x+1 by x+Ï€
x+π = 0 or x = - π
Therefore remainder = (-Ï€)3 + 3(-Ï€)2 -3(-Ï€) + 1
= - π3 + 3π2 -3π+1
Q. On dividing x3+3x2+3x+1 by 5 + 2X we get
Q. If (x−2) is a factor of x3−3x+5a then the value of a is
Ans. X – 2 = 0 or x = 2
x3−3x+5a = 0 when x = 2
(2)3−3(2)+5a = 0
8 – 6 +5a = 0 or 2+5a = 0
5a = -2 or a = -2/5 so option (d) is correct
Question. (x + 8)(x – 10) in the expanded form
is:
(a) x2 – 8x – 80
(b) x2 – 2x – 80
(c) x2 + 2x + 80
(d) x2 – 2x + 80
Ans. (x + 8)(x – 10) = x(x - 10) + 8(x - 10) = x2 - 10x + 8x – 80 = x2 - 2x - 80
Question. The value of 95 x 96 is: Ans. Correct option is B) without multiplying directly
95×96=(100−5)×(100−4) =(100)2+(−5−4)100+(−5×−4) =10000−900+20 =9120
(a) 9020
(b) 9120
(c) 9320
(d) 9340
Question. The value of 104 x 96 is: Ans. 104×96 ⇒(100+4)(100−4) ⇒(1002−42) (A2−B2=(A+B)(A−B)) ⇒10000−16 = 9984 or 1002 +(4-4)100+(4 x-4)
= 10000 =0x100 – 16 = 10000-16
(a) 9984
(b) 9624
(c) 9980
(d) 9986
Question. Without actual calculating the cubes the value of 283 + (–15)3 +(–13)3 is:
Ans. (28)3 + (−15)3 + (−13)3
∵ (28) + (-15) + (-13) = 0
Using identity, if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
= 3 × (28) × (-15) × (-13) = 16380
(a) 16380
(b) –16380
(c) 15380
(d) –15380
Question. If x – 2 is a factor of x3 – 2ax2 +ax – 1 then the value of a is: rough work
Ans. Here (x−2) is factor of f(x) ⇒f(2)=0 23−2a×22+2a−1=0 x = 2
8−8a+2a−1=0 or −6a=−7 or a=7/6
(a) 7/6
(b) –7/6
(c) 6/7
(d) 6/7
Question. If x + 2 is a factor of x3 + 2ax2 +ax – 1 then the value of a is:
Ans. Here X + 2 or X = -2 so (-2)3 + 2a(-2)2 +a(-2)
- 1 = -9 +8a-2a = -9 +6a or 6a = 9 or a=9/6
= 3/2
(a) 2/3
(b) 3/5
(c) 3/2
(d) 1/2
Question. If x + y + z = 0 then x3 + y3 + z3 is equal to
Ans. We know that x3+y3+z3–3xyz = (x+y+z)(x2+y2+z2−2xy–2yz–2zx)
If x+y+z=0 then x3+y3+z3–3xyz=0, ⟹x3+y3+z3=3xyz
(a) 3xyz
(b) – 3xyz
(c) xy
(d) –2xy
Question. The factors of 2x2 – 7x + 3 are:
Ans. Ans. 2x2-7x+3 = 2x2-6x-x+3 = 2x(x-3) -1(x-3) = (x-3)(2x-1)
(a) (x – 3)(2x – 1)
(b) (x + 3)(2x + 1)
(c) (x – 3)(2x + 1)
(d) (x + 3)(2x – 1)
Question. The factors of 6x2 + 5x – 6 are:
Ans. For the equation ax2 + bx + C, first multiply a and c, then find two factors such that their sum or difference will be equal to b.
⇒6x2+(9-4)x-6 ⇒6x2+9x-4x-6 ⇒3x(2x+3)-2(2x+3) ⇒(2x+3)(3x-2)
(a) (2x – 3)(3x – 2)
(b) (2x – 3)(3x + 2)
(c) (2x + 3)(3x – 2)
(d) (2x + 3)(3x + 2)
Question. The factors of 3x2 – x – 4 are:
Ans. 3x2-x-4 = 3x2-4x+3x-4 = x(3x-4)+1(3x-4) = (x+1)(3x-4)
(a) (3x – 4)(x – 1)
(b) (3x – 4)(x + 1)
(c) (3x + 4)(x – 1)
(d) (3x + 4)(x + 1)
Question. The factors of 12x2 – 7x + 1 are:
(b) (4x – 1)(3x + 1)
(c) (4x + 1)(3x – 1)
(d) (4x + 1)(3x + 1)
Question. The factors of x3 – 2x2 – x + 2 are:
Ans. Solution
(i) Let take f(x) = x3 - 2x2 - x + 2
The constant term in f(x) is are ±1 and ±2
Putting x = 1 in f(x), we have
f(1) = (1)3 - 2(1)2 -1 + 2
= 1 - 2 - 1 + 2 = 0
According to remainder theorem f(1) = 0 so that (x - 1) is a factor of x3
- 2x2 - x + 2
Putting x = - 1 in f(x), we have
f(-1) = (-1)3 - 2(-1)2 –(-1) + 2
= -1 - 2 + 1 + 2 = 0
According to remainder theorem f(-1) = 0 so that (x + 1) is a factor of x3
- 2x2 - x + 2
Putting x = 2 in f(x), we have
f(2) = (2)3 - 2(2)2 –(2) + 2
= 8 -82 - 2 + 2 = 0
According to remainder theorem f(2) = 0 so that (x – 2 ) is a factor of x3
- 2x2 - x + 2
Here maximum power
of x is 3 so that its can have maximum 3 factors
So our answer is (x-1)(x+1)(x-2)
(a) (x – 1)(x – 1)(x – 5)
(b) (x + 1)(x + 1)(x + 5)
(c) (x + 1)(x – 1)(x + 5)
(d) (x + 1)(x + 1)(x – 5)
Question. Which of the following is not a polynomial?
(a) x2 +√2x+3
(b) x2+ √2x+6
(c) x3+3x2+3
(d) 6x+4
Question. The degree of the polynomial 3x3 – x4 + 5x + 3 is
Ans. 3x3−x4+5x+3 =−x4+3x3+5x+3.
Then, the highest power of the variable x is 4. ∴ The degree of the given polynomial is 4.
(a) –4
(b) 4
(c) 1
(d) 3
Question. Zero of the polynomial p(x) = a2x, a≠0 is
Ans. A real number c is called a zero of the polynomial p(x) if p(c)=0.
px=0 ⇒a2x=0 ⇒x=0/a2 ⇒x=0 Hence, the correct option is option a
(a) x = 0
(b) x = 1
(c) x = –1
(d) a = 0
Question. Which of the following is a term of a
polynomial?
(a) 2x
(b) 3/x
(c) x√x
(d) x
Question. If p(x) = 5x2 – 3x + 7, then p(1) equals
Ans. Here X = 1 putting x=1
5(1)2 -3(1) + 7 = 9
(a) –10
(b) 9
(c) –9
(d) 10
Question. Factorisation of x3 + 1 is
Ans. a3+b3=(a+b)(a2−ab+b2) Hence x3+13=(x+1)(x2−(x⋅1)+12) =(x+1)(x2−x+1)
(a) (x + 1)(x2 – x + 1)
(b) (x + 1)(x2 + x + 1)
(c) (x + 1)(x2 – x – 1)
(d) (x + 1)(x2 + 1)
Question. If x + y + 2 = 0, then x3 + y3 + 8 equals
Ans. We know that a+b+c=0 ⇒a3+b3+c3=3abc.
Therefore x3+y3+23=3
× x
× y
× 2 =6xy.
(a) (x + y + 2)3
(b) 0
(c) 6xy
(d) –6xy
Question. If x = 2 is a zero of the polynomial 2x2 + 3x – p, then the value of p is
Ans. X = 2 so 2(2)2 + 3(2) – p = 8+6 = p
Therefore p=14
(a) –4
(b) 0
(c) 8
(d) 14
Question. x+1/x is
Ans. No, x+1/x= 1 is not a polynomial. A polynomial cannot have a term with the variable in the denominator, such as 1x.
(a) a polynomial of degree 1
(b) a polynomial of degree 2
(c) a polynomial of degree 3
(d) not a polynomial
Question. Integral zeroes of the polynomial (x + 3)(x – 7) are
Ans. (x+3)(x-7) = 0
X+3 =0 X-7 = 0
X=-3 X=7
so -3 and 7 is the answer
(a) –3, –7
(b) 3, 7
(c) –3, 7
(d) 3, –7
Question. The remainder when p(x) = 2x2 – x – 6 is divided by (x – 2) is
And. Here, it is given that the polynomial p(x)=2x2−x−6 and the factor is x−2, therefore, by remainder theorem, the remainder is p(2) that is: X-2 or x=2
So 2(2)2 -2 -6 = 8-8 = 0
(a) p(– 2)
(b) p(2)
(c) p(3)
(d) p(–3)
Question. If 2(a2+b2)=(a+b)2 , then
Ans.
Considering 2(a2+b2)=(a+b)2 ⇒2a2 +2b2
= a2+b2+2ab
⇒2a2−a2+2b2−b2−2ab=0
⇒a2+b2−2ab=0
⇒(a−b)2=0
⇒(a−b)=0
⇒a=b,
Hence proved.
(a) a + b = 0
(b) a = b
(c) 2a = b
(d) ab = 0
Question. If x3 + 3x2 + 3x + 1 is divided by (x + 1), then the remainder is
Ans. Let p(x) = x3 + 3x2 + 3x + 1 Here the root of x + 1 = 0 is -1 (X=-1)
p(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0
(a) –8
(b) 0
(c) 8
(d) 1/8
Question. The value of (525)2 – (475)2 is
Ans a2 – b2 =(a-b)(a+b) Here a = 525 and b = 475
So (525-475)(525+475) = 50 × 1000 = 50000
(a) 100
(b) 1000
(c) 100000
(d) –100
Question. If a + b = –1, then the value of a3 + b3 – 3ab is
Ans. Given a+b = -1 cubing on both sides
(a+b)3 = (-1)3 or a3+b3+3ab(a+b) = -1
a3+b3+3ab(-1) = -1
a3+b3-3ab =
-1
(a) –1
(b) 1
(c) 26
(d) –26
Question. The value of (2 - a)3 + (2 - b)3 + (2 - c)3 - 3(2 - a)(2 - b)(2 - c) when a + b + c = 6 is
Ans. Explanation: we know that, x+y+z=0⇔x3+y3+z3=3xyz -----(i)
Set, x=2−a,y=2−b,and,z=2−c.
Then, x+y+z=(2+2+2)−(a+b+c) =0.......[because, a+b+c=6, Given].
∴x3+y3+z3=3xyz......[∵,(i)].
⇒ The Reqd. Value=0.
(a) –3
(b) 3
(c) 0
(d) –1
17.
18 If x=1/(2−√3), then find the value of (x2−4x+1).
