# How do you factor y= x^3-2x^2+x-2?

Jun 13, 2018

$\left(x - 2\right) \left(x + i\right) \left(x - i\right)$

#### Explanation:

$\text{factor the terms "color(blue)"by grouping}$

$= \textcolor{red}{{x}^{2}} \left(x - 2\right) \textcolor{red}{+ 1} \left(x - 2\right)$

$\text{take out the "color(blue)"common factor } \left(x - 2\right)$

$= \left(x - 2\right) \left(\textcolor{red}{{x}^{2} + 1}\right)$

${x}^{2} + 1 \text{ can be factored using "color(blue)"difference of squares}$

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

$\text{with "a=x" and } b = i \to \left(i = \sqrt{-} 1\right)$

$= \left(x - 2\right) \left(x + i\right) \left(x - i\right) \leftarrow \textcolor{red}{\text{in factored form}}$

Jun 13, 2018

$y = \left(x - 2\right) \times \left({x}^{2} + 1\right)$

#### Explanation:

${x}^{3} - 2 {x}^{2}$ can be written as
${x}^{2} \times \left(x - 2\right)$, so we can re-write the original equation as

$y = {x}^{2} \times \left(x - 2\right) + \left(x - 2\right)$
or
$y = {x}^{2} \times \left(x - 2\right) + 1 \times \left(x - 2\right)$

and bringing the common factor $\left(x - 2\right)$ to the front, gives the factors:

$y = \left(x - 2\right) \times \left({x}^{2} + 1\right)$

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{x^3-2x^2+x-2 [-10, 10, -5, 5]}

Jun 13, 2018

$y = \left({x}^{2} + 1\right) \left(x - 2\right)$

#### 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.

$y = \left({x}^{2} + 1\right) \left(x - 2\right)$