CBSE Class 9 Mathematics Polynomials Worksheet Set A

Case Based MCQs

Case I : Read the following passage and answer the questions from 1 to 5.
On one day, principal of a particular school visited the classroom. Class teacher was teaching the concept of polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them.

1. Which one of the following is not a polynomial?
(a) 4x² + 2x – 1
(b) y+3/y
(c) x³ – 1
(d) y² + 5y + 1
Answer : B

2. The polynomial of the type ax² + bx + c, a = 0 is called
(a) Linear polynomial
(b) Quadratic polynomial
(c) Cubic polynomial
(d) Biquadratic polynomial
Answer : A

3. The value of k, if (x – 1) is a factor of 4x³ + 3x² – 4x + k, is
(a) 1
(b) –2
(c) –3
(d) 3
Answer : C

4. If x + 2 is the factor of x³ – 2ax² + 16, then value of a is
(a) –7
(b) 1
(c) –1
(d) 7
Answer : B

5. The number of zeroes of the polynomial x² + 4x + 2 is
(a) 1
(b) 2
(c) 3
(d) 4
Answer : B

Assertion & Reasoning Based MCQs

Directions (Q.53 to 60) : In these questions, a statement of Assertion is followed by a statement of Reason is given.

Choose the correct answer out of the following choices :
(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.

1. Assertion : 3x² + x – 1 = (x + 1) (3x – 2) + 1.
Reason : To factorise ax² + bx + c, write b as sum of two numbers whose product is ac.
Answer : A

2. Assertion : –7 is a constant polynomial.
Reason : Degree of a constant polynomial is zero.
Answer : A

3. Assertion : The degree of the polynomial (x² – 2)(x – 3)(x + 4) is 3.
Reason : A polynomial of degree 3 is called a cubic polynomial.
Answer : D

4. Assertion : If 2x² – 32 is the volume of a cuboid, then length of cuboid can be x – 8.
Reason : Volume of a cuboid = l × b × h.
Answer : D

5. Assertion : The value of 593 × 607 is 359951.
Reason : (a + b) (a – b) = a² – b²
Answer : A

6. Assertion : The value of (–27)³ + (–9)³ is –20412.
Reason : If a + b = 0, then a³ + b³ + c³ = 0
Answer : C

7. Assertion : If x = 3/2 is a zero of polynomial 2x² + kx – 12, then k = 5.
Reason : If x = a is zero of a polynomial f(x), then f(a) = 0.
Answer : A

8. Assertion : The expression 3x⁴ – 4x^3/2 + x² = 2 is not a polynomial because the term – 4x^3/2 contains a rational power of x.
Reason : The highest exponent in various terms of an algebraic expression in one variable is called its degree.
Answer : B

Question. Use suitable identities to find the following products :
(i) (x+4)(x+10)
(ii) (x+8)(x–10)
(iii) (3x+4)(3x–5)
(iv) (y²+3/2)(y²-3/2)
(v) (3–2x)(3+2x)
Answer : 
(i) (x+4)(x+10) = x²+(4+10)x+4×10
                        = x²+14x+40
(ii) (x+8)(x–10) = x²+(8−10)x+8×(−10)
                        = x²−2x−80
(iii) (3x+4)(3x–5) = 3x(3x−5)+4(3x−5)
                           = 3x×3x−3x×5+4×3x−4×5
                           = 9x²−15x+12x−20
                           = 9x²−3x−20
(iv) (y²+3/2)(y²−3/2) = (y²)² −(3/2)²
                                  = y⁴−9/4
(v) (3–2x)(3+2x) = (3)² −(2x)²
                          = 9−4x²

Question. Evaluate the following products without multiplying directly :
(i) 103×107
(ii) 95×96
(iii) 104×96
Answer : (i) 103×107 = (100+3)(100+7)
                         = (100)² +(3+7)(100)+3×7
                         = 100×100+(10)(100)+21
                         = 10000+1000+21=11021
(ii) 95×96 = (100−5)(100−4)
               = (100)² +(−5−4)(100)+(−5)(−4)
               = 100×100+(−9)(100)+20
               = 10000−900+20=9120
(iii) 104×96 = (100+4)(100−4)
                  = (100)² −(4)²
                  = 10000−16=9984

Question. Factorise the following using appropriate identifies:
(i) 9x²+6xy+y²
(ii) 4y²−4y+1
(iii) x²−y²/100
Answer : a) 9x²+6xy+y² = (3x)² +2(3x)(y)+(y)²   [(a+b)² =a²+2ab+b²]
                              = (3x+y)²
                              = (3x+y)(3x+y)
(b) 4y²−4y+1 = (2y)² −2(2y)(1)+(1)²           [(a−b)² =a²−2ab+b²]
                     = (2y−1)²
                     = (2y−1)(2y−1)
(c) x²−y²/100 = (x)2−(y/10)2                       [(x+y)(x−y)=x²−y²]
                      = (x−y/10)(x+y/10)

Question. Expand each of the following using suitable identifies:
(i) (x+2y+4z)²
(ii) (2x−y+z)²
(iii) (−2x+3y+2z)²
(iv) (3a−7b−c)²
Answer : 
(i) (x+2y+4z)²
   = x²+(2y)² +(4z)²+2(x)(2y)+2(2y)(4z)+2(4z)(x)
   = x²+4y²+16z²+4xy+16yz+8zx
(ii) (2x−y+z)²
   = [2x+(−y)+z]²
   = (2x)²+(−y)²+z²+2(2x)(−y)+2(−y)(z)+2(z)(2x)
   = 4x²+y²+z²−4xy−2yz+4zx
(iii) (−2x+3y+2z)²
   = [(−2x)+3y+2z]²
   = (−2x)²+(3y)² +(2z)² +2(−2x)(3y)+2(3y)(2z)+2(2z)(−2x)
   = 4x²+9y²+4z²−12xy+12yz−8zx
(iv) (3a−7b−c)²
    = [3a+(−7b)+(−c)]²
    = (3a)² +(−7b)²+(−c)² +2(3a)(−7b)+2(−7b)(−c)+2(−c)(3a)
    = 9a²+49b²+c²−42ab+14bc−6ca

Question. Factorise:
(i) 4x²+9y²+16z²+12xy−24yz−16xz
(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz
Answer : (i) 4x²+9y²+16z²+12xy−24yz−16xz
       = (2x)2 +(3y)² +(−4z)² +2(2x)(3y)+2(3y)(−4z)+2(2x)(−4z)
       = [2x+3y+(−4z)]²
       = (2x+3y−4z)2
(ii) 2x²+y²+8z²−2√2xy+4√2yz−8xz
    = (√2x)² +(−y)² +(−2√2z)²+2(√2x)(−y)+2(−y)(−2√2z)+2(-2√2z)(√2x)
    = [√2x+(−y)+(−2√2z)]2
   = (2√x−y−2√2z)2

Students must free download and practice these worksheets to gain more marks in exams. CBSE Class 9 Mathematics Worksheet - Polynomials