# more factoring

• Jan 5th 2008, 02:33 PM
johett
more factoring
more factoring...

a) 21$\displaystyle x^5$$\displaystyle y^4$$\displaystyle z^6$-15$\displaystyle x^4$$\displaystyle y^2$$\displaystyle z^8$+36$\displaystyle x^8$$\displaystyle y^3 z b)\displaystyle x^3-5x+4 c)4\displaystyle x^4-8\displaystyle x^3-20\displaystyle x^2-48x d)\displaystyle x^4+\displaystyle x^3-4\displaystyle x^2+5x-3 Thank you • Jan 5th 2008, 04:58 PM johett I've made some progress...not really, actually for a) i got 3\displaystyle x^4$$\displaystyle y^2$z (7x$\displaystyle y^2$$\displaystyle z^5$-5$\displaystyle z^7$+12$\displaystyle x^4$y)......I'm not sure if this is the last step or if there's more...
• Jan 5th 2008, 05:27 PM
Soroban
Hello, johett!

The first one has a common factor . . . and that's all.

The others require knowledge of the Factor Theorem.

Quote:

$\displaystyle a)\;21x^5y^4z^6 -15x^4y^2z^8+36x^8y^3z$
$\displaystyle 3x^4y^2z\left(7xy^2z^5 - 5z^7 + 12x^4y\right)$ . . . You got it! . . .Good!

Quote:

$\displaystyle b) \;x^3 - 5x+4$
$\displaystyle (x-1)(x^2+x-4)$

Quote:

$\displaystyle c)\;4x^4 -8x^3-20x^2-48x$
$\displaystyle 4x(x-4)(x^2+2x+3)$

Quote:

$\displaystyle d) \;x^4+x^3-4x^2+5x-3$
$\displaystyle (x-1)(x+3)(x^2-x+1)$