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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