Factorization-Class X ICSE Mathematics MCQs

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The solution of the MCQs of Factorization is given below 

                      Solutions 

Question 1:   When 6 x³ + 2 x² − x + 2 is divided by  (x+2) , then remainder is 

  • (a)   56  
  • (b)  36
  • (c)  44
  • (d)  − 36    

Solution :   let  f(x) = 6 x³ + 2 x² – x + 2 

                       Remainder = f(−2) = 6 (−2)³ + 2 (−2)² −(−2) +2

                                                           = 6(−8) + 2 × 4 +2+2

                                                            =  − 36 

Option (d) 

Question 2 :    When   a x³ + 6 x² +4 x + 5   is divided by (x + 3), the remainder is   −7.

  The value of constant a is

  • (a)  −2
  • (b)  2
  • (c)  3
  • (d) none of these

Solution:  Let f(x) = a x³ + 6 x² + 4 x + 5

By remainder theorem, f(−3) =  −7

⇒   a (−3)³  +  6 (−3)² + 4(−3) + 5  = −7

⇒  −27 a + 54  − 12 + 5 = −7

⇒   − 27 a =  − 54

⇒   a = 2

Option (b)

Question 3:   State whether(2x −1) is a factor of the

    polynomial  2 x³ – x² – 2 x + 1 or not 

  • (a)  yes
  • (b)  No

Yes 

Option (a)

Question 4: If  x + 1 is a factor of  3 x³ + k x² + 7 x + 4,

   then the value of k is 

  • (a)  −6
  • (b)    0
  • (c)     6
  • (d)    14

Solution:  Let f(x) = 3 x³ + k x² + 7 x +4

As   x + 1  is a factor of  f(x),  f(−1) = 0

⇒   3 (−1)³ + k (−1)² + 7 (−1) + 4 = 0

⇒    −3 + k − 7 + 4 = 0

⇒      k = 6

Option (c) 

Question 5:   If  f(x) = 2 x³ − 5 x² − p x + q  has a factor  (2 x² + x  − 1),

       then the values of p and q are

  • (a)  p = 3,  q = 4
  • (b)  p = −3,  q = 4
  • (c)  p = −4, q = 3
  • (d)  p = 4, q = 3

Solution:    2 x² + x −1 = 2 x² + 2 x – x −1 = 2x(x + 1) −(x + 1)

=  (x + 1) (2x −1)

∴ (x + 1) and (2 x − 1) is a factor of f(x)

so   f(−1) = 0

⇒ − 2 −5 + p + q = 0

⇒    p + q = 7 —— (i)

⇒    1 − 5 − 2 p + 4 q = 0

⇒  − p + 2 q = 2  ——- (ii)

Solving (i) and (ii) we get , p = 4, q = 3

Option (d)

Question 6:  If   x³ + a x² + b x  + 6  has   x − 2  as a factor and leaves

 a remainder of  3  when divided by  x − 3,  the values of a and b are

  • (a)  a = 3, b = 1
  • (b)  a = 1,  b = 3
  • (c)  a = −1, b = −3
  • (d) a = −3,  b = −1

Solution:  Let f(x) = x³ + a x² + b x  + 6 

As  x − 2  is a factor of f(x),  f(2) = 0

⇒ 2³ + a× 2² + b × 2 + 6 = 0

⇒  4 a + 2 b + 14 = 0

⇒   2a + b =  − 7    ——- (i)

according to remainder theorem, f(3) = 3

⇒ 3³ + a 3² + b × 3 + 6 = 3

⇒ 9 a + 3 b = − 30

⇒ 3a + b = − 10   ——- (ii)

Solving (i) and (ii), a = − 3  and  b = − 1

Option (d) 

Question 7:   When divided by  x −3, the polynomials 

 x³−p x² + x + 6  and 2 x³ − x² − (p+3) x − 6 leave

 the same remainder. The value of p is

  • (a)  1
  • (b)  2
  • (c)  −1
  • (d) −2

Solution:    Let  f(x) = x³−p x² + x + 6

∴  f(3) = 3³ − p × 3² + 3 + 6 = −9p + 36

Let g(x) =  2 x³ − x² − (p+3) x − 6

∴     g(3) = 2 × 3³ − 3²  −(p+3) 3 − 6  = − 3p + 30

According to question,  −9 p + 36 = − 3 p + 30

⇒  − 6 p = − 6  ⇒  p = 1

Option (a)

Question 8:  The factors of the expressions   2 x³ + 3 x² − 5 x − 6  are 

  • (a)   (x + 1) (x − 2) (2x + 3) 
  • (b)   (x + 1) (x + 2) (2 x + 3)
  • (c)   (x + 1) (x + 2) (2x – 3) 
  • (d)  (x − 1) (x − 2) (2 x + 3) 

Solution :  Let f(x) = 2 x³ + 3 x² − 5 x − 6

f(−1) = 0, so (x+1) is a factor of f(x)

Now   2 x² + x − 6 = 2 x² + 4 x − 3x − 6

= 2 x (x + 2) − 3(x + 2)

=  (x + 2) (2x − 3)

∴ f(x) = (x + 1) (x + 2) (2x − 3) 

Option (c)

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