# Factoring Imperfect squares

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• Sep 24th 2007, 01:54 PM
Freaky-Person
Factoring Imperfect squares
Well, my question is how.

Here's one from the sheet, I have the answers already, but everyone knows the answers don't matter in math >.>

1) $\displaystyle x^4 + x^2 + 25$

2) $\displaystyle 2x^4 + 8$

If you think these are both done with a pretty similar method, then you don't have to show me both. They just look different.
• Sep 24th 2007, 02:10 PM
Krizalid
Quote:

Originally Posted by Freaky-Person
1) $\displaystyle x^4 + x^2 + 25$

$\displaystyle x^4+x^2+25=(x^2+5)^2-9x^2,$ now factorise the difference of two perfect squares.
• Sep 24th 2007, 02:30 PM
topsquark
Quote:

Originally Posted by Freaky-Person
Well, my question is how.

Here's one from the sheet, I have the answers already, but everyone knows the answers don't matter in math >.>

1) $\displaystyle x^4 + x^2 + 25$

2) $\displaystyle 2x^4 + 8$

If you think these are both done with a pretty similar method, then you don't have to show me both. They just look different.

I'm not sure what you mean by "factoring imperfect squares." Neither of these expressions factor over the integers, nor do they factor over the rationals. In fact, neither of them factor over the reals, either.

By solving for the zeros of the polynomials I can tell you that
$\displaystyle x^4 + x^2 + 25 = \left ( x - \left ( \frac{3}{2} + i \cdot \frac{\sqrt{11}}{2} \right ) \right ) \left ( x - \left ( \frac{3}{2} - i \cdot \frac{\sqrt{11}}{2} \right ) \right )$$\displaystyle \left ( x - \left ( -\frac{3}{2} + i \cdot \frac{\sqrt{11}}{2} \right ) \right ) \left ( x - \left ( -\frac{3}{2} - i \cdot \frac{\sqrt{11}}{2} \right ) \right )$

You can get this by setting $\displaystyle x^4 + x^2 + 25 = 0$ and putting the factors in terms of $\displaystyle (x - r_1)(x - r_2)....$ where $\displaystyle r_1, r_2, ...$ are the zeros of the polynomial.

-Dan
• Sep 24th 2007, 02:31 PM
topsquark
Quote:

Originally Posted by Krizalid
$\displaystyle x^4+x^2+25=(x^2+5)^2-9x^2,$ now factorise the difference of two perfect squares.

Cool! Way to show me up! (Again. (Doh) )

-Dan
• Sep 24th 2007, 04:04 PM
Krizalid
Quote:

Originally Posted by Freaky-Person
2) $\displaystyle 2x^4 + 8$

$\displaystyle 2(x^4+4)$

It remains to factorise $\displaystyle x^4+4=(x^2+2)^2-4x^2$, it's the same as above.

Quote:

Originally Posted by topsquark
I'm not sure what you mean by "factoring imperfect squares."

I get it like "complete the square".

Cheers,
K.
• Sep 24th 2007, 05:17 PM
Freaky-Person
miss me?
okay, you've all been a TON of help, but then I stumbled onto a grevious problem! (oh noez)

In the following equation

$\displaystyle 4b^4 - 13b^2 + 1$

it factors to

$\displaystyle (2b^2 - 3b - 1)(2b^2 + 3b - 1)$

and I'm wondering why it's MINUS 1 and not PLUS 1
• Sep 24th 2007, 05:22 PM
WWTL@WHL
Quote:

Originally Posted by Freaky-Person
okay, you've all been a TON of help, but then I stumbled onto a grevious problem! (oh noez)

In the following equation

$\displaystyle 4b^4 - 13b^2 + 1$

it factors to

$\displaystyle (2b^2 - 3b - 1)(2b^2 + 3b - 1)$

and I'm wondering why it's MINUS 1 and not PLUS 1

Because -1 X -1 = 1.

I know 1 X 1 also = 1, but if you multiply out $\displaystyle (2b^2 - 3b + 1)(2b^2 + 3b + 1)$ it does not equal $\displaystyle 4b^4 - 13b^2 + 1$
• Sep 24th 2007, 05:28 PM
Freaky-Person
Quote:

Originally Posted by WWTL@WHL
Because -1 X -1 = 1.

I know 1 X 1 also = 1, but if you multiply out $\displaystyle (2b^2 - 3b + 1)(2b^2 + 3b + 1)$ it does not equal $\displaystyle 4b^4 - 13b^2 + 1$

I shall now try and STUMP YOU!! RAWR!!

$\displaystyle m^4 - 19m^2 + 9$

factors to

$\displaystyle (m^2 - 5m + 3)(m^2 + 5m + 3)$

LOOK!! There be a PLUS now!!

It's magic!! The wizard did it!!
• Sep 24th 2007, 05:31 PM
WWTL@WHL
I'm scared. (Sweating)
• Sep 24th 2007, 05:31 PM
Krizalid
Quote:

Originally Posted by Freaky-Person
$\displaystyle m^4 - 19m^2 + 9$

Mmm... what's the problem with this?

$\displaystyle m^4-19m^2+9=(m^2+3)^2-25m^2$, the rest is trivial.
• Sep 24th 2007, 05:39 PM
Freaky-Person
Quote:

Originally Posted by Krizalid
Mmm... what's the problem with this?

$\displaystyle m^4-19m^2+9=(m^2+3)^2-25m^2$, the rest is trivial.

look up up uuuuup!

Cause I wrote a question like this one, but it ended up with the 3rd term being NEGATIVE!! So I was like ZOMG! WHY!?!

=3
• Sep 24th 2007, 05:43 PM
WWTL@WHL
Quote:

Originally Posted by Freaky-Person
Cause I wrote a question like this one, but it ended up with the 3rd term being NEGATIVE!! So I was like ZOMG! WHY!?!

Why not? :p

What's so surprising about the 3rd term being negative? Is it the fact that you haven't seen it negative before?