# What is the standard form of  y= (x – 3)(x – 4) ?

Dec 13, 2015

$y = {x}^{2} - 8 x + 16$

#### Explanation:

Until you get used to them it seems quite tricky to multiply out brackets.

Using color to show what is happening.

Given: $y = \textcolor{b l u e}{\left(x - 3\right)} \textcolor{b r o w n}{\left(x - 4\right)}$

You can split the multiplication up into parts like this:

y=color(blue)(xcolor(brown)((x-4))-4color(brown)((x-4)) ..........(1)

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$\textcolor{b l u e}{\text{Using numbers to explain what is happening}}$
This is a bit like: $3 \times 4 = 12$
Now mater how we split the 3 we will always end up with 12

color(brown)((1+2)color(blue)(xx4)

=color(brown)((1color(blue)(xx4))+(2color(blue)(xx4))

$\textcolor{b l u e}{4 + 4 + 4} \textcolor{w h i t e}{.} = 12$

It will even work if you write 3 as 4-1

$\textcolor{b r o w n}{\left(4 - 1\right)} \textcolor{b l u e}{\times 4}$

color(brown)((4color(blue)(xx4))-(1color(blue)(xx4))  the last bracket gives you $\textcolor{g r e e n}{- 4}$

$\textcolor{b l u e}{4 + 4 + 4 + 4 \textcolor{g r e e n}{- 4}} = 12$
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$\textcolor{red}{\text{Back to your question}}$

So equation (1) becomes:

$y = \left({x}^{2} - 4 x\right) \textcolor{red}{-} \left(\textcolor{g r e e n}{+ 4 x - 16}\right) \ldots \ldots \ldots \left(2\right)$

As there is no multiplication left we can removing the brackets

$y = {x}^{2} - 4 x \textcolor{g r e e n}{\textcolor{b l u e}{\boldsymbol{-}} 4 x \textcolor{b l u e}{+} 16} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right)$

Notice the way the $\textcolor{red}{\text{minus sign")color(white)(.) underline("outside}}$ the $\underline{\text{last}}$ bracket in equation (2) changes all the $\textcolor{b l u e}{\text{signs inside}}$ it when the bracket is removed. Giving equation (3)

$y = {x}^{2} - 8 x + 16$