Show that (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24.

Let f(x) = x³ – 3x² – 10x + 24

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 – 2) to be a factor, we will find f(2)

⇒ f(2) = (2)³ – 3(2)² – 10(2) + 24

⇒ f(2) = 8 – 12 – 20 + 24

⇒ f(2) = 0

So, (x–2) is a factor.

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

⇒ f(–3) = (–3)³ – 3(–3)² – 10(–3) + 24

⇒ f(–3) = –27 – 27 + 30 + 24

⇒ f(–3) = 0

So, (x+3) is a factor.

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

⇒ f(4) = (4)³ – 3(4)² – 10(4) + 24

⇒ f(4) = 64 – 48 – 40 + 24

⇒ f(4) = 0

So, (x–4) is a factor.

∴ (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24