Author Topic: Someone help solve this cubic equation (Read 835 times) Tweet Share

Jsy443 · « **on:** March 22, 2018, 10:05:48 am »

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

Someone explain to me how to figure this equation out, will appreciate it. Thank you.

jazzycab · « **Reply #1 on:** March 22, 2018, 11:58:28 am »

Quote from: Jsy443 on March 22, 2018, 10:05:48 am

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

Someone explain to me how to figure this equation out, will appreciate it. Thank you.

By the rational root theorem, the only possible rational roots are -2, -1, 1 or 2.
We can check each one using the remainder theorem:
$\begin{align*}\text{Let }P\left(x\right)&=x^3-x^2-2x+2\\P\left(-2\right)&=-8-4+4+2\\&=-6\\\therefore x+2&\text{ and }-x-2\text{ are not factors of }P\left(x\right)\\P\left(-1\right)&=-1-1+2+2\\&=2\\\therefore x+1&\text{ and }-x-1\text{ are not factors of }P\left(x\right)\\P\left(1\right)&=1-1-2+2\\&=0\\\text{Thus }x-1&\text{ is a factor of }P\left(x\right)\end{align*}$
A cubic with a known linear factor must also have a quadratic factor. Let the quadratic factor be $ax^2+bx+c$:
$\begin{align*}P\left(x\right)&=x^3-x^2-2x+2\\&=\left(x-1\right)\left(ax^2+bx+c\right)\\&=ax^3+\left(b-a\right)x^2+\left(c-b\right)x-c\\\therefore ax^3&=x^3\\\left(b-a\right)x^2&=-x^2\\\left(c-b\right)x=-2x\\-c&=2\\\therefore a&=1\\b-1&=-1\\c&=-2\\-2-b&=-2\\b&=0\\\therefore P\left(x\right)&=\left(x-1\right)\left(x^2-2\right)\\&=\left(x-1\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\end{align*}$

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Author Topic: Someone help solve this cubic equation (Read 835 times) Tweet Share

Jsy443

Someone help solve this cubic equation

jazzycab

Re: Someone help solve this cubic equation

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