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# What the zeros of the polynomial function ...?

### 2 Answers

- llafferLv 72 months ago
The same logic as your last question is done here. You need to find any rational roots so you can factor the polynomial down to a quadratic to find the last two roots.

The Rational Root Theorem says that if there are any rational roots it will be from a list made up of all factors of the constant term over all factors of the high-degree polynomial. In this case, that list is:

±1, ±2, ±4, ±8, ±16

There are 10 possible rational roots that we'd have to brute-force to try to find one.

Using the multiple-choice answers as a hint, we see it should be either -2 or +2. So I'll test the twos and look for zeroes:

f(x) = x³ + 5x² - 6x - 16

f(-2) = (-2)³ + 5(-2)² - 6(-2) - 16 and f(2) = 2³ + 5(2)² - 6(2) - 16

f(-2) = -8 + 5(4) + 12 - 16 and f(2) = 8 + 5(4) - 12 - 16

f(-2) = -8 + 20 + 12 - 16 and f(2) = 8 + 20 - 12 - 16

f(-2) = 8 and f(2) = 0

We now know that x = 2 is a root which means that (x - 2) is a factor. This means we can divide the cubic by this factor to get a quadratic quotient:

. . . . _x²_+_7x_+_8___

x - 2 ) x³ + 5x² - 6x - 16

. . . . . x³ - 2x²

. . . . -------------

. . . . . . . . 7x² - 6x - 16

. . . . . . . . 7x² - 14x

. . . . . . . ---------------

. . . . . . . . . . . . . 8x - 16

. . . . . . . . . . . . . 8x - 16

. . . . . . . . . . . . ------------

. . . . . . . . . . . . . . . . . 0

Now that we have the resulting quadratic to get the final two roots:

x² + 7x + 8 = 0

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -7 ± √(7² - 4(1)(8))] / (2 * 1)

x = [ -7 ± √(49 - 32)] / 2

x = (-7 ± √17) / 2

The three roots are:

x = 2 and (-7 ± √17) / 2

Answer D.

- ?Lv 72 months ago
f(x) = x^3 + 5x^2 - 6x - 16

f(x) = (x - 2) (x^2 + 7 x + 8)

x = 2

x = - 7/2 - sqrt(17)/2

x = - 7/2 + sqrt(17)/2

Answer choice:

D