Question

If x + 3 and 2x + 1 are the factors of 6x³ + px² + qx − 6, then the value of p is ______.

Options

• −5

• 5

• 17

• −17

Answer 1

If x + 3 and 2x + 1 are the factors of 6x³ + px² + qx − 6, then the value of p is 17.

Explanation:

Let f(x) = 6x³ + px² + qx − 6

∵ (x + 3) is a factor of f(x)

x + 3 = 0

x = −3

∴ f (−3) = 0

⇒ 6(−3)³ + p(−3)² + q(−3) − 6 = 0

⇒ 6(−27) + p(−9) + q(−3) − 6 = 0

⇒ −162 + 9p − 3q − 6 = 0

⇒ 9p − 3q − 168 = 0

⇒ 9p − 3q = 168

Divided by 3,

3p − q = 56   ...(i)

∵ (2x + 1) is a factor of f(x)

2x + 1 = 0

x = `-1/2`

∴ `f (-1/2)` = 0

⇒ `6(-1/2)^3 + p(-1/2)^2 + q(-1/2) − 6` = 0

⇒ `-6/8 + p/4 - q/2 - 6` = 0

Multiplying the entire equation by 4,

⇒ − 3 + p − 2q − 24 = 0

⇒ p − 2q = 27   ...(ii)

From Equation (i), we can express q in terms of p:

q = 3p − 56

Substitute this expression for q into Equation (ii):

p − 2(3p − 56) = 27

p − 6p + 112 = 27

−5p = 27 − 112

−5p = −85

p = `(-85)/(-5)`

p = 17