One root of f(x) = x³ - 4x² - 20x + 48 is x = 6. What are all the factors of the function? Use the Remainder Theorem.
(x + 6)(x + 8)
(x - 6)(x - 8)
(x - 2)(x + 4)(x - 6)
(x + 2)(x - 4)(x + 6)

Solution:

f(x) = x³ - 4x² - 20x + 48

If x = 6 is one root then using remainder theorem

Substitute x = 6 we get

f (6) = (6)³ - 4(6)² - 20(6) + 48

By further calculation

f (6) = 216 - 4 (36) - 120 + 48

f (6) = 216 - 144 - 120 + 48

f (6) = 0

Using the synthetic division, the quotient is x² + 2x - 8

Now split the middle term

x² + 4x - 2x - 8

Taking out the common terms

x (x + 4) - 2 (x + 4)

So we get

(x + 4) (x - 2)

Therefore, the other factors are (x + 4) and (x - 2).

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One root of f(x) = x³ - 4x² - 20x + 48 is x = 6. What are all the factors of the function? Use the Remainder Theorem.
(x + 6)(x + 8)
(x - 6)(x - 8)
(x - 2)(x + 4)(x - 6)
(x + 2)(x - 4)(x + 6)

Summary:

One root of f(x) = x³ - 4x² - 20x + 48 is x = 6. All the factors of the function are (x + 4), (x - 2) and (x-6).