Question

5. Find the values of constants a and b when
x- 2 and x + 3 both are the factors of
expression x3 + ax2 + bx - 12.
​

Answer 1

Answer :-

→ a = 3 and b = -4 .

Step-by-step explanation :-

We have,

An Expression :- x³ + ax² + bx – 12 .

$\because$ (x – 2) is a factor i.e., x = 2 the remainder will be zero .

⇒ (2)³ + a(2)² + b(2) – 12 = 0 .

⇒ 8 + 4a + 2b – 12 = 0 .

⇒ 4a + 2b = 4 .

⇒ 2a + b = 2 . ...….(i)

$\because$ When x + 3 is a factor i.e., x = - 3 the remainder will be zero.

⇒ (- 3)³ + a(- 3)² + b(- 3) – 12 = 0 .

⇒ - 27 + 9a – 3b – 12 = 0 .

⇒ 9a – 3b = 39 .

⇒ 3a – b = 13. ......(ii)

On adding equation (i) and (ii) , we get

2a + b = 2 .

3a - b = 13.

+....-........+

_____________

⇒ 5a = 15 .

$\therefore$ a = 3 .

Putting the value of a in the equation (i), we get

⇒ 2 × 3 + b = 2.

⇒ 6 + b = 2 .

⇒ b = 2 – 6 .

$\therefore$ b = -4 .

Hence, a = 3 and b = - 4 .

Answer 2

Answer:

The answer is in the attachment. Pls prefer attachment