1. ## OMG... poly factoring?

Ok... I am soooooo lost with this.

I need to factor out to find zeros of an equation.

–x4 + 10x2 + 8x – 8 is the equation.

I don't understand.... AT ALL... how they factored out the (x + 2) to get this:

(x + 2)( $-x^3+2x^2+6x-4$)

I understand "factoring out" something, but I don't understand how to factor out say... x+2

Thanks guys... this is really bothering me, and I have a test tomorrow. Definitely need to understand this concept.

3. I know how to do that stuff. Wait... since it's x-2 or x+2 (Since x is a power of 1) I can just do synthetic division? Or should I use long division for that?

I can't believe this section is about synthetic and long division and I never bothered to think to use it. lol

4. ahaha Yes your absolutely right.

so you would do this. since synthetic divison is (x-k) (notice that minus), and so what would happen is

for (x+2)

-2| -1 0 10 8 -8
-1 2 6 -4 0
-1 2 6 -4 0

Since remainder is 0, x-k is a factor, (x+2).

The remaining thing at bottom (-1 2 6 -4) would translate into -x^3 + 2x^2 + 6x -4. Get it

5. Awesome... however, this question still has me stumped.

Can anyone explain how the equation:

$6x^4 - 31x^3 - 32x^2 + 11x + 6$

has the zeros: -1, -1/3, 1/2, and 6?

I found -1 and 6 myself... but how does one check if -1/3 is a zero? You clearly can't do synthetic with fractions, and long division seems WAY too messy.

Thanks guys

6. of course you can do synthetic with fractions

1/3 6 -31 -32 11 6

Try it out

7. Ah! I see what I did wrong.

However.... if you're checking for zeros and are NOT allowed to use a calculator, what would the best method be?

Thanks again, btw. This stuff just boggled my mind because I missed a day of class. :/

ALSO!!!!!!!! Is the only time you break the equation down and then do the quadratic on it when you check EVERY zero and only 2 of them work? Cause then you clearly have to find the other 2 (Assuming it's $x^4$)?

8. synthetic division.

to your question, well you can keep breaking them down, its just easier to check the broken down version than the long veriosn. Besides, if only two work, then you'll end with a quadratic, then you're forced to find the other two (remember that they can be imaginary roots)

9. ## ahhh

I got most of it figured out!

HOWEVER...

$\frac{-2 \pm \sqrt{4-4(1)(-4)}}{2}$

is.... $\frac{-2 \pm \sqrt{20}}{2}$

How do I break that down to get the final ones? An example in the book was like... $\sqrt{8}$ and they turned that into 2 * $\sqrt{2}$

10. cause like since
$
\sqrt{8}
$

is
$
\sqrt{2*4}
$

then its
$
\sqrt{2}*\sqrt{4}
$

$
2\sqrt{2}
$

11. I always forget silly rules like this.

Thanks man, you've been a BIGGGGGGGGG help!

12. no prob man. just need for my questions to be answered now ;(

13. –x^4 + 10x^2 + 8x – 8 = 0, factoring

-(x^4 - 10x^2 - 8x + 8) = 0, the method of factoring is by group, add or subtrct something to complete the factoring,

-[x^4 + 2x^3 - 2x^3 - 10x^2 - 8x + 8] = 0, split -10x^2

-[x^4 + 2x^3 - 2x^3 - 4x^2 - 6x^2 - 8x + 8] = 0, re-arrange,

-[(x^4 + 2x^3)- 2x^3 - 4x^2- 6x^2 - 8x + 8] = 0, add and subtract -4x,

-[(x^4 + 2x^3)- 2x^3 - 4x^2- 6x^2 - 8x - 4x + 4x + 8] = 0, grouping,

-[(x^4 + 2x^3)+(-2x^3 - 4x^2)+(-6x^2 - 8x +-4x) + (4x + 8)] = 0,

factor again the common factor for each parenthesis,

-[x^3(x + 2) - 2x^2(x + 2) - 6x(x + 2) + 4(x + 2)] = 0, factor (x + 2),

-(x + 2)(x^3 - 2x^2 - 6x + 4) = 0, regroup again, add/subtract 2x^2 and also 2x,

-(x + 2)(x^3 - 2x^2 + 2x^2 - 2x^2 - 6x - 2x + 2x + 4) = 0,

-(x + 2)(x^3 + 2x^2 - 4x^2 - 6x - 2x + 2x + 4) = 0, factor again,

-(x + 2)(x^3 + 2x^2 - 4x^2 - 8x + 2x + 4) = 0,

-(x + 2)(x^2(x + 2) - 4x( x + 2)+ 2(x + 2)= 0,

-(x + 2)(x + 2)(x^2 - 4x + 2) = 0,

using quadratic formula for (x^2 - 4x + 2) = 0, the factoring becomes

-(x + 2)(x + 2)(x - (2 + sqrt 2))(x -(2 - sqrt 2)) = 0,

roots are: -2, -2, (2 + sqrt 2), (2 - sqrt 2).

see plot,

-