# Thread: Factoring polynomials with exponents

1. ## Factoring polynomials with exponents

I'm currently stuck on 2 problems.
$4(3-x)^2-(3-x)^3+3(3-x)$ I got $3-x(6(3-x)(3-x)^2)$
That doesn't seem right but what went wrong.

also

$15(2z+1)^3+10(2z+1)^2-25(2z+1)$ => $5(2z+1)(3(2z+1)^2+2(2z+1)-5)$ => $5(2z+1)((2z+1)^2(2z+1))$
which also doesn't seem right.

2. Originally Posted by m0nkeysensei
I'm currently stuck on 2 problems.
$4(3-x)^2-(3-x)^3+3(3-x)$ I got $3-x(6(3-x)(3-x)^2)$
That doesn't seem right but what went wrong.

also

$15(2z+1)^3+10(2z+1)^2-25(2z+1)$ => $5(2z+1)(3(2z+1)^2+2(2z+1)-5)$ => $5(2z+1)((2z+1)^2(2z+1))$
which also doesn't seem right.
check your steps ,you are doing mistake somewhere

check spoiler if stuck
Spoiler:

$4 { \color{blue}(3-x)^2 }- { \color{blue} (3-x)^3}+3 { \color{blue} (3-x)}={ \color{blue} (3-x)} \{ 4{ \color{blue} (3-x) }- { \color{blue} (3-x)^2 }+3 \}$
solving
$= (3-x) \{ 4(3-x)- (9+x^2-6x)+3 \}$
$= (3-x) ( 12-4x- 9-x^2+6x+3)$
$=(3-x) (-x^2+2x+6)$
$= - (3-x) (x-1+ \sqrt {7})(x-1-\sqrt {7}) =(x-3)(x-1+ \sqrt {7})(x-1-\sqrt {7})$

3. Originally Posted by ramiee2010

Spoiler:

$4 { \color{blue}(3-x)^2 }- { \color{blue} (3-x)^3}+3 { \color{blue} (3-x)}={ \color{blue} (3-x)} \{ 4{ \color{blue} (3-x) }- { \color{blue} (3-x)^2 }+{ \color{blue} (3-x)} \}$
solving
$= (3-x) \{ 4(3-x)- (9+x^2-6x)+(3-x) \}$
$= (3-x) ( 12-4x- 9-x^2+6x)+3-x )$
$=(3-x) (-x^2+2x+6)$
$= - (3-x) (x-1+ \sqrt {7})(x+1+\sqrt {7}) =(x-3)(x-1+ \sqrt {7})(x+1+\sqrt {7})$
but how does
if you are taking out a 3-x from all terms, wouldn't that cancel out the last 3-x?

4. Originally Posted by m0nkeysensei
but how does
if you are taking out a 3-x from all terms, wouldn't that cancel out the last 3-x?
that was mistyped by me. i have been corrected that