# Determining x-intercepts of polynomial function

• March 8th 2011, 11:08 AM
Devi09
Determining x-intercepts of polynomial function
How do you determine the x-intercepts of the polynomial function y = -(2x+5)^3 - 20

I did this:

0 = -(2x+5)^3 - 20

20 = -2x+5)^3

-5 3sqrt(20/-2) (3 before the sqrt being the little 3)

x= 10.7

The answer should be -3.86 however.
• March 8th 2011, 11:28 AM
masters
Quote:

Originally Posted by Devi09
How do you determine the x-intercepts of the polynomial function y = -(2x+5)^3 - 20

I did this:

0 = -(2x+5)^3 - 20

20 = -2x+5)^3

-5 3sqrt(20/-2) (3 before the sqrt being the little 3)

x= 10.7

The answer should be -3.86 however.

Hi Devi09,

See if you can follow this:

$-(2x+5)^3-20=0$

$-(2x+5)^3=20$

$-(2x+5)=\sqrt[3]{20}$

$-2x-5=\sqrt[3]{20}$

$-2x=\sqrt[3]{20}+5$

$x=\dfrac{\sqrt[3]{20}+5}{-2}$

• March 9th 2011, 02:34 AM
HallsofIvy
Quote:

Originally Posted by Devi09
How do you determine the x-intercepts of the polynomial function y = -(2x+5)^3 - 20

I did this:

0 = -(2x+5)^3 - 20

20 = -2x+5)^3

You've dropped a "(". It should be 20= -(2x+5)^3. Now "undo" that. -20^{1/3}= 2x+ 5
2x= -20^{1/3}- 5, x= (-20^{1/3})/2

Quote:

-5 3sqrt(20/-2) (3 before the sqrt being the little 3)
"cube root" or "cbrt" not "square root" nor "3sqrt"
But your real error is that you divided by two before taking the cuberoot. You must take the cuberoot first because the "2x+ 5" is inside the cube.

Quote:

x= 10.7

The answer should be -3.86 however.