How do I factor x^7+2x^6-2x^5-4x^4+x^3+2x^2 ?
\[\large x^7+2x^6-2x^5-4x^4+x^3+2x^2 \] First notice you can pull out a common factor of x^2: \[\large x^2(x^5+2x^4-2x^3-4x^2+x+2)\] Now we want to find one factor of that, so we can use synthetic division. Hint: try checking x=1.
Oh, yes, I did that, but I needed to find the roots as well. I saw on Wolfram Alpha, that it's supposed to end up looking like \[x^2(x-1)(x-1)(x+1)(x+1)(x+2)=0\], but it doesn't explain exactly how to get that particular answer... Hmmm...
Do you know how to do synthetic division?
Dividing out the factor (x - 1) by synthetic division yields: \[x^2(x-1)(x^4+3x^3+x^2-3x-2)\] Notice if you put x=1 into the parentheses, you'll get zero again. So divide again by x=1, to pull out another (x-1) factor.
If you don't know how to do synthetic division, you might want to google it to see how it's done. Once you get it, it's easy.
I've done synthetic division before. I'm taking a Calculus class right now. I just don't see why it wasn't pointed out in my previous reference. I'll study your answer. Thank You for the help!
Just keep dividing out factors until you're left with a quadratic, which you can then factor easily. You can use a couple of methods to find factors eg rational roots theorem.
Okay. I see where it's headed. The (x-1) threw me off. Albeit, I'm still confused. How did that monomial get selected in the first place, from this whole equation? \[x^7+2x^6-2x^5-4x^2+x+1\]?
Whoops. Correction to the previous equation. It should look like this instead: \[x^7+2x^6-2x^5-4x^4+x^3+2x^2=0\].
I chose x=1 because if you put x=1 into that equation, it equals zero... so (x-1) has to be a factor. Same thing with the next x=1 factor.
Okay. I'll take note of that. Thanks again!
(x-1) is often a factor in these types of questions, so it's good to check x=1 in the equation first.
Ah, I see...
nevermind idiots.
k.
@Fireflame13: You spelled "nevermind" wrong.
did i spell ugly wrong?
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