# If f(-2) = 0, what are all the factors of the function f(x) = x^{3} - 2x^{2 }- 68x - 120. Use the Remainder Theorem

(x + 2)(x + 60)

(x - 2)(x - 60)

(x - 10)(x + 2)(x + 6)

(x + 10)(x - 2)(x - 6)

**Solution:**

Given f(-2) = 0, clearly (x + 2) is one of the factors of f(x) = x^{3} - 2x^{2 }- 68x - 120.

Using remainders theorem,

f(x) = q(x)(x + 2) = r(x)

Since (x + 2) is a factor, r(x) = 0. Now to find the remaining factor i.e.,

p(x) = f(x)/(x + 2)

Using long division method,

Therefore, the quotient is x^{2 }- 4x - 60.

To find remaining factors,

x^{2 }- 4x - 60 =0

x^{2} - 10x + 6x - 60 = 0

x(x - 10) + 6(x - 10) = 0

(x + 6)(x - 10) = 0

Therefore, the factors of the given function are (x + 2), (x + 6) and (x - 10).

## If f(-2) = 0, what are all the factors of the function f(x) = x^{3} - 2x^{2 }- 68x - 120. Use the Remainder Theorem

**Summary:**

The factors of the function f(x) = x^{3} - 2x^{2 }- 68x - 120 are (x + 2), (x + 6) and (x - 10).

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