Question

Find the value of a such that (x-4) is a factor of $5x^3-7x^2$ - ax - 28

Answer 1

Let f(x) $5x^3-7x^2$ - ax - 28

and Let x - 4 = 0, then x = 4

Now, f(4) $=5(4{)}^3-7(4{)}^2-a\times$ 4 - 28

= $5\times 64-7\times$ 16 - 4a - 28

= 320 - 112 - 4a - 28

= 320 - 140 - 4a = 180 - 4a

$\therefore$ (x-4) is a factor of f(x)

$\therefore$ Remainder = 0

$\Rightarrow$ 180 - 4a = 0 $\Rightarrow$ 4a = 180

$\Rightarrow a=\frac{180}{4}=45$

$\therefore$ a = 45