Show that (x + 4), (x – 3) and (x – 7) are factors of x³ – 6x² – 19x + 84.

Let f(x) = x³ – 6x² – 19x + 84

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0

For checking (x + 4) to be a factor, we will find f(–4)

⇒ f(–4) = (–4)³ – 6(–4)² – 19(–4) + 84

⇒ f(–4) = –64 – 96 + 76 + 84

⇒ f(–4) = 0

So, (x+4) is a factor.

For checking (x – 3) to be a factor, we will find f(3)

⇒ f(3) = (3)³ – 6(3)² – 19(3) + 84

⇒ f(3) = 27 – 54 – 57 + 84

⇒ f(3) = 0

So, (x–3) is a factor.

For checking (x – 7) to be a factor, we will find f(7)

⇒ f(7) = (7)³ – 6(7)² – 19(7) + 84

⇒ f(7) = 343 – 294 – 133 + 84

⇒ f(7) = 0

So, (x–7) is a factor.

∴ (x + 4), (x – 3) and (x – 7) are factors of x³ – 3x² – 10x + 24