Solve x^2(a^2 + 2ab + b^2) - x(a + b) = 0 by factoring

vanbeersj

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I need to solve by factoring: x²(a²+2ab+b²) - x(a+b) = 0.
I have so far factored it to x²(a+b)(a+b) - x(a+b) = 0, I know I can factor this further based on the (a+b), I think it's x²-x + (a+b)(a+b) = 0, but it doesn't make sense. Please help, I'm confused as to my next step.
 
vanbeersj said:
I need to solve by factoring: x²(a²+2ab+b²) - x(a+b) = 0.
I have so far factored it to x²(a+b)(a+b) - x(a+b) = 0....
Excellent! You've actually done the hard part already. :wink:

Now reformat a little:

. . . . .x[sup:2jjinck9]2[/sup:2jjinck9](a + b)[sup:2jjinck9]2[/sup:2jjinck9] - x(a + b) = 0

. . . . .[ x(a + b) ][sup:2jjinck9]2[/sup:2jjinck9] - [ x(a + b)] = 0

. . . . .[ x(a + b) ][sup:2jjinck9]2[/sup:2jjinck9] - 1[ x(a + b) ] = 0

You can factor and solve (2y)[sup:2jjinck9]2[/sup:2jjinck9] - 2y = 0. Apply the exact same reasoning here, and you're done! :D

Eliz.
 
Re: Quadratic Equation

Hello, vanbeersj!

Solve by factoring: .\(\displaystyle x^2(a^2+2ab+b^2) - x(a+b) \;= \;0\)

I have so far factored it to: .\(\displaystyle x^2(a+b)(a+b) - x(a+b) \:= \:0\)
I know I can factor this further based on the \(\displaystyle (a+b)\) . . . . and the x

\(\displaystyle \text{We have: }\;\underbrace{x^2}(a+b)\underbrace{(a+b)} - \underbrace{x}\underbrace{(a+b)} \:=\:0\)

\(\displaystyle \text{Both terms have a common }x\text{ and a common }(a+b)\)

\(\displaystyle \text{Factor them out: }\;x(a+b)\cdot\bigg[x(a+b) - 1\bigg] \:=\:0\)


\(\displaystyle \text{You know the routine: set each factor equal to zero and solve.}\)

. . \(\displaystyle x(a+b) \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:0}\)

. . \(\displaystyle x(a+b) - 1 \:=\:0 \quad\Rightarrow\quad x(a+b) \:=\:1 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{a+b}}\)

 
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