# Factoring

• Jul 5th 2011, 04:35 PM
theloser
Factoring
Factor
To go about the first one please show me the steps. Thanks
1. $(x+5)^4(x-9)^3+(x+5)^4(x-9)^4$
$(x+5)^4(x-9)^3[(1)+(x-9)]$?

2. $x^4-15x+144$
Prime?
• Jul 5th 2011, 04:39 PM
pickslides
Re: Factoring
Quote:

Originally Posted by theloser
Factor
To go about the first one please show me the steps. Thanks
1. $(x+5)^4(x-9)^3+(x+5)^4(x-9)^4$
$(x+5)^4(x-9)^3[(1)+(x-9)]$?

This is correct so far, to finish

$(x+5)^4(x-9)^3[(1)+(x-9)]$

$(x+5)^4(x-9)^3(x-8)]$

Quote:

Originally Posted by theloser
2. $x^4-15x+144$
Prime?

This one looks tricky, are you sure its not $x^4-15x^2+144$ ??
• Jul 5th 2011, 04:44 PM
theloser
Re: Factoring
Quote:

Originally Posted by pickslides
This is correct so far, to finish

$(x+5)^4(x-9)^3[(1)+(x-9)]$

$(x+5)^4(x-9)^3(x-8)]$

This one looks tricky, are you sure its not $x^4-15x^2+144$ ??

O.O I made a mistake...
It is $x^4-15x^2+144$
its not a perfect squared and I don't see any factors of 144 that would total -15.
• Jul 5th 2011, 04:48 PM
Also sprach Zarathustra
Re: Factoring
Quote:

2. $x^4-15x+144$
Prime?
[/QUOTE]

$x^4>15x-144$ for all x in R ===> no factorization over R.
• Jul 5th 2011, 04:49 PM
Also sprach Zarathustra
Re: Factoring
Quote:

Originally Posted by theloser
O.O I made a mistake...
It is $x^4-15x^2+144$
its not a perfect squared and I don't see any factors of 144 that would total -15.

Substitute:

x^2=t
• Jul 5th 2011, 05:22 PM
theloser
Re: Factoring
I think its prime.
• Jul 5th 2011, 05:31 PM
Quacky
Re: Factoring
See the graph.

Not that you needed confirmation, but the discriminant (once you've made Zach's substitution) would have told you at a glance.