# How do you simplify (x-3)^2-4(x-3)+3?

Apr 13, 2018

$\left(x - 6\right) \left(x - 4\right)$

#### Explanation:

${\left(x - 3\right)}^{2} - 4 \left(x - 3\right) + 3$

$\left(\left(x - 3\right) - 3\right) \left(\left(x - 3\right) - 1\right)$

$\left(x - 6\right) \left(x - 4\right)$

Apr 16, 2018

Simplified gives ${x}^{2} - 10 x + 24$

and factorised give: $\left(x - 6\right) \left(x - 4\right)$

#### Explanation:

To simplify we need to multiply out the brackets:

${\left(x - 3\right)}^{2} - 4 \left(x - 3\right) + 3$

$= {x}^{2} - 6 x + 9 - 4 x + 12 + 3$

$= {x}^{2} - 10 x + 24$

However, I suspect that the question was meant to be to factorise..

We could multiply out as done above and then factorise from this, but let's regard $\left(x - 3\right)$ as a variable, $p$.

${\left(x - 3\right)}^{2} - 4 \left(x - 3\right) + 3$

$\rightarrow {p}^{2} - 4 p + 3$

$= \left(p - 3\right) \left(p - 1\right)$

$\rightarrow \left(x - 3 - 3\right) \left(x - 3 - 1\right)$

$= \left(x - 6\right) \left(x - 4\right)$