Factoring Question

**astuart** · September 24th 2012, 10:38 PM

Hello,

I can't figure out how I should factor this expression.

$x^4 - 18x^2 + 1$

I know the answer is $(x^2 - 4x - 1)(x^2 + 4x - 1)$

But I don't know how it gets to this point. I've tried substituting x squared with y to get this...

$y^2 - 18y + 1$

but this can't really be factored much more, so I don't know what I should be doing..

**Prove It** · September 24th 2012, 11:10 PM

Originally Posted by astuart

Hello,

I can't figure out how I should factor this expression.

$x^4 - 18x^2 + 1$

I know the answer is $(x^2 - 4x - 1)(x^2 + 4x - 1)$

But I don't know how it gets to this point. I've tried substituting x squared with y to get this...

$y^2 - 18y + 1$

but this can't really be factored much more, so I don't know what I should be doing..

You could complete the square...

**Soroban** · September 25th 2012, 08:22 AM

Hello, astuart!

This require complete-the-square,
. . but in a different manner.

$\text{Factor: }\:x^4 - 18x^2 + 1$

We have: . $x^4 + 1 - 18x^2$

Subtract and add $2x^2\!:\;x^4\; {\color{red}-\; 2x^2} + 1 - 18x^2 \;{\color{red}+\; 2x^2}$

. . . . . . . . . . . . . . $=\;(x^4 - 2x^2 + 1) - 16x^2$

. . . . . . . . . . . . . . $=\;(x^2-1)^2 - (4x)^2$

. . . . . . . . . . . . . . $=\;\left([x^2-1)-4x\right]\left([x^2-1]+4x\right)$

. . . . . . . . . . . . . . $=\;(x^2-4x-1)(x^2+4x-1)$

**astuart** · September 25th 2012, 01:55 PM

Originally Posted by Soroban

Hello, astuart!

This require complete-the-square,
. . but in a different manner.

We have: . $x^4 + 1 - 18x^2$

Subtract and add $2x^2\!:\;x^4\; {\color{red}-\; 2x^2} + 1 - 18x^2 \;{\color{red}+\; 2x^2}$

. . . . . . . . . . . . . . $=\;(x^4 - 2x^2 + 1) - 16x^2$

. . . . . . . . . . . . . . $=\;(x^2-1)^2 - (4x)^2$

. . . . . . . . . . . . . . $=\;\left([x^2-1)-4x\right]\left([x^2-1]+4x\right)$

. . . . . . . . . . . . . . $=\;(x^2-4x-1)(x^2+4x-1)$

Ah, so by subtracting 2x^2 from the equation, it results in 16x^2, which is a perfect square.
I was confused by the subtraction of the 2x^2, I didn't know what the reason for it was..

Thanks

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