Find a zero of f(x) and factor out the associated term given the following:

Question

f(x)=2x³-21x²+57x-18

Answer 1

f(x) = 2x³ - 21x² + 57x - 18

I use the synthetic division to find a zero that will give me a remainder of zero based on the remainder theorem.

These are the possible solutions: ±1, ±2, ±3, ±6, ±9, and ±18.

Turns out 6 will work because of two things:

1) f(6) = 2(6)³ - 21(6)² + 57(6) - 18 = 432 - 756 + 342 - 18 = 0

2) 6| 2 -21 57 -18

^↓ 12 -54 18

2 -9 3 |0|

(x - 6)(2x² - 9x + 3)

If finding solutions, then we will use the quadratic formula to find the zeros for 2x² - 9x + 3.

x = [-b ± √(b² - 4ac)] ÷ 2a

x = [-(-9) ± √((-9)² - 4(2)(3))] ÷ 2(2)

x = [9 ± √(81 - 24)] ÷ 4

x = [9 ± √(57)] ÷ 4

x = [0.363 , 4.137]

In this case, there are 3 solutions. x = 0.363, 4.137, and 6.

Answer 2

Hi there!

f(x) = 2x³ - 21x² + 57x - 18

Looking at the function, I can see that there's a greatest common factor between the last three values : 6.

To verify, I will try plugging that value in.

f(6) = 2(6)³ - 21(6)² + 57(6) - 18 = 0

This means that x = 6 is a factor, so we'll factor out (x-6).

f(x) = (x - 6) (2x² - 9x + 3) = 0