Question

Let ƒ(x) = x6 – 40x4 – 40x3 – 141x2 – 200x + 420. Sergei claims that x + 2 is a factor of f(x). To prove his claim, what must he show to be the value of ƒ(– 2)? ƒ(– 2) =

Answer 1

f(-2) = 0

Explanation:

If x+2 is a factor of f(x), then -2 will be a zero of f(x).

Sergei must show that f(-2) = 0.

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Proof:

f(x) = x^6 – 40x^4 – 40x^3 – 141x^2 – 200x + 420.

f(-2) = (-2)⁶ - 40(-2)⁴ - 40(-2)³ - 141(-2)² - 200(-2) + 420

f(-2) = 64 – 40(16) - 40(-8) – 141(4) – 200(-2) + 420

f(-2) = 64 – 640 + 320 – 564 +400 + 420

f(-2) = 0

The graph of f(x) also shows that (x+ 2) is a factor, because there is an

x-intercept at (-2, 0).