# need help on some factor problems

• Nov 30th 2006, 10:31 PM
elfen_lied
need help on some factor problems
can someone factor these problems until they cant be factored anymore thanks:)

a^3+a^2b-ab^2-b^3

x^4-10x^2+9

x^4-24x^2+144

a^3+2a^2-5a-10
• Nov 30th 2006, 10:52 PM
CaptainBlack
Quote:

Originally Posted by elfen_lied
can someone factor these problems until they cant be factored anymore thanks:)

a^3+a^2b-ab^2-b^3

\$\displaystyle a^3+a^2b-ab^2-b^3=a^2(a+b)-b^2(a+b)=(a^2-b^2)(a+b)\$\$\displaystyle =(a-b)(a+b)(a+b)\$

RonL
• Nov 30th 2006, 11:01 PM
CaptainBlack
Quote:

Originally Posted by elfen_lied
can someone factor these problems until they cant be factored anymore thanks:)
x^4-10x^2+9

Clearly if \$\displaystyle x=1\$ this is zero so \$\displaystyle x-1\$ is a factor. Long division gives:

\$\displaystyle
x^4-10x^2+9=(x-1)(x^3+x^2-9x-9)
\$

and if \$\displaystyle x=-1\$ the second term on the right is zero, so \$\displaystyle x+1\$ is a factor
of that term, and you should be able to complete this yourself from this
point.

RonL
• Nov 30th 2006, 11:29 PM
elfen_lied
Quote:

Originally Posted by CaptainBlack
Clearly if \$\displaystyle x=1\$ this is zero so \$\displaystyle x-1\$ is a factor. Long division gives:

\$\displaystyle
x^4-10x^2+9=(x-1)(x^3+x^2-9x-9)
\$

and if \$\displaystyle x=-1\$ the second term on the right is zero, so \$\displaystyle x+1\$ is a factor
of that term, and you should be able to complete this yourself from this
point.

RonL

i don't quite get what you just said there can you explain it a bit more?
is (x-1) the GCF?
• Nov 30th 2006, 11:32 PM
CaptainBlack
Quote:

Originally Posted by elfen_lied
i dont quite get what you just said there can you explain it a bit more?

if a polynomial p(x) is zero when x=a, then x-a is a factor of the polynomial.

When x=-1, x^3+x^2-9x-9=0, so x+1 is a factor of x^3+x^2-9x-9.

RonL
• Nov 30th 2006, 11:46 PM
elfen_lied
Quote:

Originally Posted by CaptainBlack
if a polynomial p(x) is zero when x=a, then x-a is a factor of the polynomial.

When x=-1, x^3+x^2-9x-9=0, so x+1 is a factor of x^3+x^2-9x-9.

RonL

can you please just factor the problems step by step because i really don't understand anything in words only in numbers:)
• Nov 30th 2006, 11:58 PM
CaptainBlack
Quote:

Originally Posted by elfen_lied
can you please just factor the problems step by step because i really don't understand anything in words only in numbers:)

Why don't you post your work and then we will be able to tell you where you are going wrong.

You will not benefit from us doing all your work for you, we are here to help you learn to do it for yourself.

RonL
• Dec 1st 2006, 12:06 AM
elfen_lied
Quote:

Originally Posted by CaptainBlack
Why don't you post your work and then we will be able to tell you where you are going wrong.

You will not benefit from us doing all your work for you, we are here to help you learn to do it for yourself.

RonL

okay so after (x-1)(x^3+x^2-9x-9)
so do i factor them some more?
if so how?
• Dec 1st 2006, 03:47 AM
topsquark
Quote:

Originally Posted by elfen_lied
x^4-10x^2+9

Perhaps a slightly different approach would work better for you.

\$\displaystyle x^4-10x^2+9\$

Notice how we only have even powers of x listed here. So to momentarily simplify the expression I'm going to define \$\displaystyle y = x^2\$:
\$\displaystyle x^4-10x^2+9 = y^2 - 10y + 9\$

This is a simple quadratic, which you should know how to factor.

I get that
\$\displaystyle y^2 - 10y + 9 = (y - 1)(y - 9)\$

Now we put the x's back in:
\$\displaystyle x^4-10x^2+9 = y^2 - 10y + 9 = (y - 1)(y - 9) = (x^2 - 1)(x^2 - 9)\$

\$\displaystyle x^4-10x^2+9 = (x^2 - 1)(x^2 - 9)\$

Now. Notice that each of these factors is of the form: \$\displaystyle x^2 - a^2 = (x + a)(x - a)\$, so

\$\displaystyle x^4-10x^2+9 = (x^2 - 1)(x^2 - 9) = (x + 1)(x - 1)(x + 3)(x - 3)\$

The same approach will also work on your third problem.

-Dan