# Questions about factoring an equation with 3rd degree and 2nd degree

• April 7th 2011, 08:58 AM
sara213
Questions about factoring an equation with 3rd degree and 2nd degree
$2x^3+3x^2-1$

This is the equation that needs to be fully factored. So, from what I see here it would be impossible to factor it normally, since 1 can't be multiplied by 3 or any other number to get 1 other then 1.

I figure this is a good idea to use factor theorem. I figured out that p(-1) =0 that would mean that x+1 is a factor of this equation.

next I did division

Attachment 21393

I got stuck half way through the division when I reached the (x^2 -1)(x^2-x) I don't think i can subtract 1-x. Not sure where to go from here. (Worried) any help appreciated.
• April 7th 2011, 09:32 AM
veileen
"I figured out that p(-1) = 0 that would mean that x+1 is a factor of this equation." Good, so you have: $2x^3+3x^2-1=2x^3+2x^2+x^2+x-x-1=(x+1)(2x^2+x-1)$.
I don't understand what you did in that picture, but from here, in general $ax^2+bx+c=0$ can be factored: $a(x-x_1)(x-x_2)$.

$(x^2-1)(x^2-x)$ ~ $a^2-b^2=(a+b)(a-b)$, for the second bracket x is common factor.
• April 7th 2011, 09:36 AM
TheEmptySet
Quote:

Originally Posted by sara213
$2x^3+3x^2-1$

This is the equation that needs to be fully factored. So, from what I see here it would be impossible to factor it normally, since 1 can't be multiplied by 3 or any other number to get 1 other then 1.

I figure this is a good idea to use factor theorem. I figured out that p(-1) =0 that would mean that x+1 is a factor of this equation.

next I did division

Attachment 21393

I got stuck half way through the division when I reached the (x^2 -1)(x^2-x) I don't think i can subtract 1-x. Not sure where to go from here. (Worried) any help appreciated.

Your idea is sound but you need to remember to include all powers of x when you do division

To clarify you started with

$2x^3+3x^2-1=2x^3+3x^2+0x-1$

$\begin{array}{crrrrrrr}
& 2x^2&+& x &-&1 &&\\ \cline{2-8}
(x+1) \vline & 2x^3 &+& 3x^2&+&0&-&1 \\
&-(2x^3 & + & 2x^2) & &&&\\
&&& x^2 & + & 0x &&\\
&&&-(x^2 & + & x) &&\\
&&&&&-x & - & 1 \\
&&&&&-(-x & - & 1) \\
&&&&&&&0

\end{array}$
• April 7th 2011, 09:57 AM
sara213
Quote:

Originally Posted by TheEmptySet
Your idea is sound but you need to remember to include all powers of x when you do division

To clarify you started with

$2x^3+3x^2-1=2x^3+3x^2+0x-1$

$\begin{array}{crrrrrrr}
& 2x^2&+& x &-&1 &&\\ \cline{2-8}
(x+1) \vline & 2x^3 &+& 3x^2&+&0&-&1 \\
&-(2x^3 & + & 2x^2) & &&&\\
&&& x^2 & + & 0x &&\\
&&&-(x^2 & + & x) &&\\
&&&&&-x & - & 1 \\
&&&&&-(-x & - & 1) \\
&&&&&&&0

\end{array}$

Sorry about the picture...I tired to post it form math type into here and it became huge so I did a gif but that didn't work out well. I wanted to show the division in full process but I couldn't figure out how to add the spaces.

• April 7th 2011, 10:08 AM
tonio
Quote:

Originally Posted by sara213
$2x^3+3x^2-1$

This is the equation that needs to be fully factored. So, from what I see here it would be impossible to factor it normally, since 1 can't be multiplied by 3 or any other number to get 1 other then 1.

I figure this is a good idea to use factor theorem. I figured out that p(-1) =0 that would mean that x+1 is a factor of this equation.

next I did division

Attachment 21393

I got stuck half way through the division when I reached the (x^2 -1)(x^2-x) I don't think i can subtract 1-x. Not sure where to go from here. (Worried) any help appreciated.

Besides long division you can do this by simple "eye", meaning: just complete the second factor to what you need:

$2x^3+3x^2-1=(x+1)(2x^2+x-1)$ ...how did I find the factor $2x^2+x-1$ ? Well, we need $2x^3$ , so we must

multiply $x\mbox { by } 2x^2$ . After this we need $3x^2$ , and we already have $(x+1)(2x^2...)$ , so we have so

far only $2x^2$ , so we need to add $x$ ...and etc.

Of course, after this we can continue $2x^2+x-1=(x+1)(2x-1)$ and etc.

Tonio