Question:
Find the value of a such that $(x-4)$ is a factor of $5 x^{3}-7 x^{2}-a x-28$
Solution:
Here, $f(x)=5 x^{3}-7 x^{2}-a x-28$
By factor theorem
If (x - 4) is the factor of f(x) then, f(4) = 0
⟹ x – 4 = 0
⟹ x = 4
Substitute the value of x in f(x)
$f(4)=5(4)^{3}-7(4)^{2}-a(4)-28$
= 5(64) – 7(16) – 4a – 28
= 320 – 112 – 4a – 28
= 180 – 4
Equate f(4) to zero, to find a
f(4) = 0
⟹ 180 – 4a = 0
⟹ -4a = -180
⟹ 4a = 180
⟹ a = 180/4
⟹ a = 45
When a = 45, (x - 4) will be factor of f(x)