pre calc question

Question

4x^3-9x-14

Find all zeros of f(x)=4x3-9x-14. Enter the zeros separated by commas. Enter exact value, not decimal approximations.

Answer 1

I'm going to assume that you cannot use a calculator for this.

According to Descartes rule of signs, since there is 1 sign change, there will be 1 positive real root. So, I'll start guessing positive real roots that comply with the Rational Zeros (Roots) Theorem. The Rational Zeros Theorem says that IF there are any rational zeros, they will be te factors of 14 divided by the factors of 4 so: 14, 7, 7/2, 2, 7/4, 1, or 1/2. I'll start doing synthetic division on the integers and star with "1" first:

1 | 4 0 -9 -14

| 4 4 -5

|___________

4 4 -5 -19 (did not produce a zero)

Try 2:

2 | 4 0 -9 -14

| 8 16 14

|___________

4 8 7 0 (This WORKED!)

2 is one of the zeros. The remaining quadratic was also found. It is 4x² + 8x + 7

Now that you have a quadratic, you can use the quadratic formula to find the other zeros:

x = [-b ± √(b² - 4ac)]/(2a)

a = 4, b = 8 c = 7:

x = [-8 ± √(8² - 4•4•7)]/(2•4)

x = [-8 ± √(64 - 112)]/8

x = [-8 ± √(-48)]/8

x = [-8 ± √(16•3•(-1))]/8

x = [-8 ± (4√3)i]/8

x = [-2 ± (√3)i]/2

So the 2nd zero is (-2 + √3 i)/2 and the 3rd zero is (-2 - √3 i)/2

Answer 2

f(x) = 4x³ - 9x - 14

By trial and error, we see that f(2) = 0. So, x - 2 is a factor.

Divide f(x) by x - 2 to get f(x) = (x - 2)(4x² + 8x + 7)

Use the quadratic formula to find the remaining roots. (They are imaginary)