Factoring Imperfect squares

**Freaky-Person** · September 24th 2007, 01:54 PM

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) $x^4 + x^2 + 25$

2) $2x^4 + 8$

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.

**Krizalid** · September 24th 2007, 02:10 PM

Originally Posted by Freaky-Person

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

$x^4+x^2+25=(x^2+5)^2-9x^2,$ now factorise the difference of two perfect squares.

**topsquark** · September 24th 2007, 02:30 PM

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) $x^4 + x^2 + 25$

2) $2x^4 + 8$

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
$x^4 + x^2 + 25 = \left ( x - \left ( \frac{3}{2} + i \cdot \frac{\sqrt{11}}{2} \right ) \right ) \left ( x - \left ( \frac{3}{2} - i \cdot \frac{\sqrt{11}}{2} \right ) \right )$ $\left ( x - \left ( -\frac{3}{2} + i \cdot \frac{\sqrt{11}}{2} \right ) \right ) \left ( x - \left ( -\frac{3}{2} - i \cdot \frac{\sqrt{11}}{2} \right ) \right )$

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

-Dan

**topsquark** · September 24th 2007, 02:31 PM

Originally Posted by Krizalid

$x^4+x^2+25=(x^2+5)^2-9x^2,$ now factorise the difference of two perfect squares.

Cool! Way to show me up! (Again.

)

-Dan

**Krizalid** · September 24th 2007, 04:04 PM

Originally Posted by Freaky-Person

2) $2x^4 + 8$

$2(x^4+4)$

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

Originally Posted by topsquark

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

I get it like "complete the square".

Cheers,
K.

**Freaky-Person** · September 24th 2007, 05:17 PM

okay, you've all been a TON of help, but then I stumbled onto a grevious problem! (oh noez)

In the following equation

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

it factors to

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

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

**WWTL@WHL** · September 24th 2007, 05:22 PM

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

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

it factors to

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

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 $(2b^2 - 3b + 1)(2b^2 + 3b + 1)$ it does not equal $4b^4 - 13b^2 + 1$

**Freaky-Person** · September 24th 2007, 05:28 PM

Originally Posted by WWTL@WHL

Because -1 X -1 = 1.

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

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

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

factors to

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

LOOK!! There be a PLUS now!!

It's magic!! The wizard did it!!

**WWTL@WHL** · September 24th 2007, 05:31 PM

I'm scared.

**Krizalid** · September 24th 2007, 05:31 PM

Originally Posted by Freaky-Person

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

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

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

**Freaky-Person** · September 24th 2007, 05:39 PM

Originally Posted by Krizalid

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

$m^4-19m^2+9=(m^2+3)^2-25m^2$ , 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

**WWTL@WHL** · September 24th 2007, 05:43 PM

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?

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

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