MCQ for Class 9 Mathematics Polynomials

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) – π/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-π

Ans. P(x) = 5x – π

            Zero’s of p(x), we get p(x) = 0  

            i.e. 5x – π = 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+π

            Let p(x) = x³+3x²+3x+1 by x+π

            x+π = 0 or x = - π

            Therefore remainder = (-π)³ + 3(-π)² -3(-π) + 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:

(a) (4x – 1)(3x – 1)
(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).