# Factoring

Printable View

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

2.\$\displaystyle 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.\$\displaystyle (x+5)^4(x-9)^3+(x+5)^4(x-9)^4\$
\$\displaystyle (x+5)^4(x-9)^3[(1)+(x-9)]\$?

This is correct so far, to finish

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

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

Quote:

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

This one looks tricky, are you sure its not \$\displaystyle 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

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

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

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

O.O I made a mistake...
It is \$\displaystyle 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.\$\displaystyle x^4-15x+144\$
Prime?
[/QUOTE]

\$\displaystyle 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 \$\displaystyle 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.