Carol Valentine

2021-12-17

How do you factor $y={x}^{3}-2{x}^{2}+x-2$?

alexandrebaud43

$\left(x-2\right)\left(x+i\right)\left(x-i\right)$
Explanation:
factor the terms by grouping
$={x}^{2}\left(x-2\right)+1\left(x-2\right)$
take out the common factor $\left(x-2\right)$
$=\left(x-2\right)\left({x}^{2}+1\right)$
${x}^{2}+1$ can be factored using difference of squares
${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$
with $a=x$ and $b=i⇒\left(i=\sqrt{-1}\right)$
$=\left(x-2\right)\left(x+i\right)\left(x-i\right)⇒\text{in factored form}$

alexandrebaud43

$y=\left(x-2\right)×\left({x}^{2}+1\right)$
Explanation:
${x}^{3}-2{x}^{2}$ can be written as ${x}^{2}×\left(x-2\right)$
$y={x}^{2}×\left(x-2\right)+\left(x-2\right)$
or
$y={x}^{2}×\left(x-2\right)+1×\left(x-2\right)$
and bringing the common factor $\left(x-2\right)$ to the front, gives the factors:
$y=\left(x-2\right)×\left({x}^{2}+1\right)$
solving the equation for $y=0$ gives the solution $x=2$ meaning that the graph of the equation crosses the x-axis at the point (2,0)
solving the equation for $x=0$ gives $y=-2$ , meaning that the graph of the equation crosses the y-axis at (0,-2)
graph $\left\{{x}^{3}-2{x}^{2}+x-2\left[-10,10,-5,5\right]\right\}$

RizerMix

$y=\left({x}^{2}+1\right)\left(x-2\right)$
Explanation:
$y={x}^{3}-2{x}^{2}+x-2$
We will use a factoring method called grouping. If we group the first and the third and the second and the fourth together, each grout has its own greatest common factor:
$y=x\left({x}^{2}+1\right)-2\left({x}^{2}+1\right)$
Now we see another CF between the two terms : ${x}^{2}+1$ . We can factor it out.
$y=\left({x}^{2}+1\right)\left(x-2\right)$

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