Question

If 3x^4 + 6x^3– ax^2 – bx – 12 is completely divisible by x^2– 3, then find the values of a and b.​

Answer 1

Answer:

a = 5 ; b = 18

See the attachment for answer.

Answer 2

Answer:

a = 5, b = 18

Step-by-step explanation:

$f(x)=3x^{4} + 6x^{3} -ax^{2} -bx-12$

If f(x) is completely divisible by x^2– 3 = (x - √3)(x + √3) ⇒ that ±√3 are zeroes of given polynomial.

f(√3) = $3(\sqrt{3})^{4} +6(\sqrt{3})^{3} -a(\sqrt{3})^{2} -b\sqrt{3} -12$ = 0

27 + 18√3 - 3a - b√3 - 12 = 0

3a + b√3 = 15 + 18√3 ..... (1)

f(-√3) = $3(-\sqrt{3}^{4} + 6(-\sqrt{3} )^{3} - a(-\sqrt{3})^{2} -b(-\sqrt{3}) -12 = 0$

27 - 18√3 - 3a + b√3 - 12 = 0

3a - b√3 = 15 - 18√3 ..... (2)

(1) + (2)

6a = 30 ⇒ a = 5

(1) - (2)

2b√3 = 36√3 ⇒ b = 18

$f(x)=3x^{4} +6x^{3} -5x^{2} -18x-12$