Question

$x-3$ and $x+3$ are the factors of $4x^3+a{x}^2+bx$. Find the values of a and b.

• A. a = 0, b = 36
• B. a = 0, b = -36
• C. a = 5, b = -36
• D. a = 8, b = 36

Answer 1

The correct option is B a = 0, b = -36

Let $f\left(x\right)=4x^3+a{x}^2+bx$
(x-3) and (x+3) are factors of f(x)
By factor theorem,
f(3) = 0
$\Rightarrow 4(3{)}^3+a(3{)}^2+b\left(3\right)=0$
$\Rightarrow 108+9a+3b=0$
$\Rightarrow 36+3a+b=0$
$\therefore b=-3a-36$ .........(1)
f(-3) = 0
$\Rightarrow 4(-3{)}^3+a(-3{)}^2+b(-3)=0$
$\Rightarrow -108+9a-3b=0$
$\Rightarrow -36+3a-b=0$
$\therefore b=-36+3a$ .........(2)

From (1) and (2), we get
-3a - 36 = -36 +3a
$\Rightarrow -6a=0$
$a=0$
from (1), we get:
b = -36 - 3a = -36 - 0
$\therefore b=-36$

So, the correct answer is option (b).