Find the value of a such that (x - 4) is a factor of

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)