# How do you find the product of (x-3)^2?

Apr 4, 2018

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

From here we can just FOIL; first, outer, inner, last. Multiply the first term in the first bracket (x), by the FIRST term in the second bracket (x), giving us ${x}^{2}$. Then, we multiply the outermost term on the left side by the OUTERmost on the right side, giving us $- 3 \cdot x$, equal to $- 3 x$. We then multiply the INNERmost terms together, which are again -3 and $x$, giving us $- 3 x$ again. Finally we multiply the LAST term in the first bracket by the last term of the second, giving us $- 3 \cdot - 3$, or 9. Then, we simply write these out, in order:

${x}^{2} - 3 x - 3 x + 9$

Collect like terms:

${x}^{2} - 6 x + 9$

And you're done!

Apr 4, 2018

${x}^{3} - 6 x + 9$

#### Explanation:

1. Let's say $A = x - 3$ (this is to help you understand).
So then we would have ${A}^{2}$, which gets you $A \times A$.

2. Now let's apply that back into the original problem.
${\left(x - 3\right)}^{2} = \left(x - 3\right) \times \left(x - 3\right) = \left(x - 3\right) \left(x - 3\right)$

3. We can use the Distributive Property to make this more easily readable:
$\setminus \textcolor{red}{x \left(x - 3\right)} + \setminus \textcolor{g r e e n}{\left(- 3\right) \left(x - 3\right)}$
Simplify by re-applying Distributive Property:
$\setminus \textcolor{red}{x \left(x\right) + x \left(- 3\right)} + \setminus \textcolor{g r e e n}{\left(- 3\right) \left(x\right) + \left(- 3\right) \left(- 3\right)}$

4. And multiply...
${x}^{2} + \left(- 3 x\right) + \left(- 3 x\right) + 9 = {x}^{3} - 6 x + 9$

$\setminus \textcolor{\to m a \to}{\setminus \textrm{\left(F \mathmr{and} t h e t h i r d s t e p\right\rangle}}$ This is really just me explaining FOIL in a different way.