# factoring a higher order trinomial (odd)

• Dec 15th 2012, 07:40 PM
slowlyGettingIt
factoring a higher order trinomial (odd)
Hi - I am stuck on a question that requires me to find the 2nd point of intersection between a power function (x^3) and the tangent of the function at a given point (1/2, 1/8), which I determined to be y=3/4x - 1/4

I have combined the two equations and created a 3rd order trinomial 0=x^3 -3/4x +1/4. My problem is that I forget how to factor a trinomial with an odd power. Any help would be appreciated.

• Dec 15th 2012, 08:09 PM
jakncoke
Re: factoring a higher order trinomial (odd)
Honestly, its really hard to remember the cubic one,

but if you need it solved, click the link.

0&#61;x&#94;3 -3&#47;4x &#43;1&#47;4 - Wolfram|Alpha
• Dec 15th 2012, 08:45 PM
slowlyGettingIt
Re: factoring a higher order trinomial (odd)
Thanks for the link - I see it being of value in the future. However, I really want to figure this one out.
• Dec 16th 2012, 05:22 AM
a tutor
Re: factoring a higher order trinomial (odd)
But you know that this line is a tangent to the cubic so you have a repeated root at x=1/2.

Now you just need to find the other factor.
• Dec 16th 2012, 11:03 AM
slowlyGettingIt
Re: factoring a higher order trinomial (odd)
Thanks a tutor. I understand that the first point of intersection is 1/2 and that the tangent will intersect at a second point, which can be determined by figuring out the second factor, but that is exactly my problem. I cannot remember how to determine the other factor for a cubic trinomial that cannot be simplified into a quadratic. I believe mine cannot 0=x^3 -(3/4)x + 1/4.

Honestly, it has been a while since I studied these earlier concepts and I used to struggle with factoring. Is there a different theorem or method that can be applied?
• Dec 16th 2012, 11:18 AM
skeeter
Re: factoring a higher order trinomial (odd)
Quote:

0=x^3 -(3/4)x + 1/4
$\displaystyle x = \frac{1}{2}$ is a zero, use synthetic division ...

Code:

[1/2].........1.........0........-3/4..........1/4
.......................1/2........1/4..........-1/4
-------------------------------------------------------
..............1........1/2.......-1/2............0

$\displaystyle x^3 - \frac{3}{4}x + \frac{1}{4} = \left(x - \frac{1}{2}\right)\left(x^2 + \frac{1}{2}x - \frac{1}{2}\right)$
• Dec 16th 2012, 11:51 AM
a tutor
Re: factoring a higher order trinomial (odd)
I was thinking

$\displaystyle x^3-\frac{3}{4}x+\frac{1}{4}=(x-1/2)^2(?+?)$

It's pretty clear that it has to be $\displaystyle x^3-\frac{3}{4}x+\frac{1}{4}=(x-1/2)^2(x+1)$

I think I'm lazy.
• Dec 16th 2012, 12:35 PM
slowlyGettingIt
Re: factoring a higher order trinomial (odd)
Thanks guys. You both have helped a lot.