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

alexandrebaud43

Beginner2021-12-19Added 36 answers

$y=(x-2)\times ({x}^{2}+1)$
Explanation:
$x}^{3}-2{x}^{2$ can be written as ${x}^{2}\times (x-2)$ $y={x}^{2}\times (x-2)+(x-2)$
or
$y={x}^{2}\times (x-2)+1\times (x-2)$
and bringing the common factor $(x-2)$ to the front, gives the factors:
$y=(x-2)\times ({x}^{2}+1)$
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 $\{{x}^{3}-2{x}^{2}+x-2[-10,10,-5,5]\}$

RizerMix

Skilled2021-12-29Added 437 answers

$y=({x}^{2}+1)(x-2)$
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: $y=x({x}^{2}+1)-2({x}^{2}+1)$
Now we see another CF between the two terms : ${x}^{2}+1$ . We can factor it out. $y=({x}^{2}+1)(x-2)$