# And now what?

• Nov 13th 2008, 02:22 AM
Alienis Back
And 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 AM
clic-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 AM
Alienis 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 PM
mr fantastic
Quote:

Originally Posted by Alienis Back
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.

$\displaystyle 16x^2+16x-1$

$\displaystyle 8x(2x+2)-1$ Mr F says: This will not help.

:confused:

Note that $\displaystyle 16x^2 + 16x - 1 = (4x+2)^2 - 5 = (4x + 2)^2 - (\sqrt{5})^2$.

Now use the difference of two squares formula.
• Nov 17th 2008, 02:42 AM
Alienis 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 AM
mr fantastic
Quote:

Originally Posted by Alienis 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?

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 AM
Alienis 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 AM
mr fantastic
Quote:

Originally Posted by Alienis Back
[snip]
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.

It's just my way of completing the square. You look like you've got your equally good way of completing the square which works well for you. Neither way is better of worse than the other. Whatever floats your boat .....