I have to factor the following polynomial completely. But I don't have the answer so I don't know if I did it or it can be factored further.

$\displaystyle

16x^2+16x-1

$

$\displaystyle

8x(2x+2)-1

$

:confused:

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- Nov 13th 2008, 02:22 AMAlienis BackAnd now what?
I have to factor the following polynomial completely. But I don't have the answer so I don't know if I did it or it can be factored further.

$\displaystyle

16x^2+16x-1

$

$\displaystyle

8x(2x+2)-1

$

:confused: - Nov 13th 2008, 10:02 AMclic-clac
Finding this polynomial roots will allow you to factor it completely.

So the problem is to find the roots.

Try before looking at the solution (which can be wrong):

$\displaystyle 16x^{2}+16x-1=16(x+\frac{1+\sqrt{\frac{5}{4}}}{2})(x+\frac{1-\sqrt{\frac{5}{4}}}{2})$ - Nov 14th 2008, 09:53 AMAlienis Back
I think the answer has to be a lot easier, the book i'm using hasn't aproached roots so far. I'm just factoring polynomials. Said that, I don't understand the point of findig the roots. I looked at your explanation and...well, maybe the book author meant "factor completly" according to what we have covered so far...Anyway, I'll keep trying.

I notice the second part looks very much like a difference of squares.

$\displaystyle

a^2-b^2=(a+b)(a-b)

$

...And how do I find roots if that were the only chance? Prease an explanation - Nov 14th 2008, 03:10 PMmr fantastic
- Nov 17th 2008, 02:42 AMAlienis Back
Ok! This works for me but how did you know that $\displaystyle 16x^2 + 16x - 1$ would be equivalent to $\displaystyle (4x+2)^2 - 5$

I mean...as it is not a perfect square trinomial I couldn't try finding two factors adding to the product of the external values as I did with this:

$\displaystyle

16x^5 - 44x^4 +30x^3

$

$\displaystyle

2x^3 (8x^2 - 22x + 15)

$

I usually find the numerical product of the external values:

$\displaystyle

8(15)=120

$

And then try to find two factors of 120 that add to the middle term. i.e.$\displaystyle

-22

$

$\displaystyle

-10(-12)=120

$

Finally all I have to do is just replacing:

$\displaystyle

2x^3 (8x^2-10x-12x+15)

$

And solve...

$\displaystyle

2x^3 [2x(4x-5)-3(4x-5)]

$

$\displaystyle

2x^3 (4x-5) (2x-3)

$

And that's it!

As you see this aproach wouldn't have worked with $\displaystyle 16x^2 + 16x - 1$ as it isn't a perfect square. Now my question is how can I know $\displaystyle (4x+2)^2 - 5$ will solve for the above trinomial? Is there any special formula or technique? - Nov 17th 2008, 03:16 AMmr fantastic
I got the expression by completing the square:

$\displaystyle 16x^2 + 16x - 1 = 16\left(x^2 + x - \frac{1}{16}\right)$

$\displaystyle = 16 \left( \left[x + \frac{1}{2}\right]^2 - \frac{1}{4} - \frac{1}{16}\right)$

$\displaystyle = 16\left(x + \frac{1}{2}\right)^2 - 4 - 1$

$\displaystyle = 4^2 \left( x + \frac{1}{2} \right)^2 - 5$

$\displaystyle = \left( 4x + 4 \cdot \frac{1}{2} \right)^2 - 5$

$\displaystyle = \left( 4x + 2 \right)^2 - 5$.

Now compare this with the difference of two squares formula $\displaystyle A^2 - B^2 = (A - B)(A + B)$:

Substitute $\displaystyle A = 4x + 2$ and $\displaystyle B = \sqrt{5}$. - Nov 18th 2008, 03:21 AMAlienis Back
Then here is how it works. You factored out the $\displaystyle 16$:

$\displaystyle 16x^2 + 16x - 1 = 16\left(x^2 + x - \frac{1}{16}\right)$

$\displaystyle = 16 \left[ \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{1}{16}\right]$

__________________________________________________ ________________

But I didn't know where you obtained that $\displaystyle \left(x+\frac{1}{2}\right)^2$ so I solved...

$\displaystyle (x+\frac{1}{2}) (x+\frac{1}{2})$

$\displaystyle =x^2+\frac{x}{2}+\frac{x}{2}+\frac{1}{4}$

$\displaystyle =x^2+x+\frac{1}{4}$

And replaced...

$\displaystyle =16 [x^2+x+\frac{1}{4}-\frac{1}{4}-\frac{1}{16}]$

$\displaystyle =16\left[x^2+x-\frac{1}{16}\right]$

__________________________________________________ ________________

So now I continue...

$\displaystyle =16\left[\left(x+\frac{1}{2}\right)^2-\frac{1}{4}-\frac{1}{16}\right]$

$\displaystyle =16\left[\left(x+\frac{1}{2}\right)^2-\frac{4}{16}-\frac{1}{16}\right]$

and the $\displaystyle 16$ outside the parentesis will multiply for $\displaystyle \frac{4}{16}$ and $\displaystyle \frac{1}{16}$ so we now have...

$\displaystyle =16\left(x+\frac{1}{2}\right)^2-4-1$

$\displaystyle =4^2\left(x+\frac{1}{2}\right)^2-4-1$

$\displaystyle =4^2\left(x+\frac{1}{2}\right)^2-5$

Consider... $\displaystyle (a^n) (b^n) = (ab)^n$

$\displaystyle =(4x+\frac{4}{2}\right)^2 - 5$

$\displaystyle =(4x+2)^2-5$

$\displaystyle =(4x+2)^2-(\sqrt{5})^2$

And finally we apply the formula for the difference of two squares...

$\displaystyle =[(4x+2)-\sqrt{5}] [(4x+2)+\sqrt{5}]$

And that's it.

Now...there is just one little thing that I don't understand. Hope you don't think I'm a pain in the... but how did you get that $\displaystyle -\frac{1}{4}$ up there on the second stage?. I mean...It seems you just put it there arbitrarily so the whole thing could be worked out. Finally I want to thank you. Sincerely. - Nov 18th 2008, 04:00 AMmr fantastic