# 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

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