One factor of f(x) = 5x³ + 5x² - 170x + 280 is (x + 7). What are all the roots of the function? Use the remainder theorem.

Solution:

f(x) = 5x³ + 5x² - 170x + 280

One factor is x + 7

It can be written as x = - 7

Using the remainder theorem, we know that if (x+7) is a factor of f(x), then the remainder of f(-7) is 0

f(7) = 5(-7)³ + 5(-7)² - 170(-7) + 280

= 5(-343)+5(49) +1190+280

=0

Thus f(-7) = 0

If we divide f(x) by x + 7 we get 5x² - 30x + 40.

Thus 5x³ + 5x² - 170x + 280 = (x + 7)(5x² - 30x + 40)

5x² - 30x + 40 = 0

Let us take 5 as common

5(x² - 6x + 8) = 0

By factoring this quadratic equation by splitting the middle terms, we get

x² - 2x - 4x + 8 = 0

By taking out x as common in the first two terms and 4 as common in the next two terms

x (x - 2) - 4 (x - 2) = 0

(x - 4) (x - 2) = 0

x - 4 = 0 and x - 2 = 0

x = 4 and x = 2

Therefore, the roots of the function are x = -7, 4 or 2.

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One factor of f(x) = 5x³ + 5x² - 170x + 280 is (x + 7). What are all the roots of the function? Use the remainder theorem.

Summary:

One factor of f(x) = 5x³ + 5x² - 170x + 280 is (x + 7). The roots of the function are x = -7, 4 or 2.