Question

One factor of f(x)5x^3+5x^2-170+280 is (x + 7). What are all the roots of the function? Use the Remainder Theorem.


a. x = –4, x = –2, or x = 7

b. x = –7, x = 2, or x = 4

c. x = –7, x = 5, or x = 280

d. x = –280, x = –5, or x = 7

Answer 1

Answer: (b) x = -7, x = 2, x = 4

Step-by-step explanation:

Remainder Theorem is used to determine if a given value is a root.

It is stated that (x + 7) is a root ⇒ x + 7 = 0  ⇒ x = -7

We can confirm this by plugging in x = -7 and getting a value of 0.

f(x) = 5x³ + 5x² - 170x + 280

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

      = -1715 + 245 + 1190 + 280

     = 0

CONFIRMED that x = -7 is a zero!

Next, let's try x = 2

f(2) = 5(2)³ + 5(2)² - 170(2) + 280

      = 40 + 20 - 340 + 280

     = 0

CONFIRMED that x = 2 is a zero!

Lastly, let's try x = 4

f(4) = 5(4)³ + 5(4)² - 170(4) + 280

      = 320 + 80 - 680 + 280

     = 0

CONFIRMED that x = 4 is a zero!