# What is the standard form of  y= (4x-15)(2x-2)-(3x-1)^2?

Apr 7, 2018

$y = - {x}^{2} - 32 x + 29$

Here's how I did it:

#### Explanation:

Standard form means that we have to put the equation in this form: $y = a {x}^{2} + b x + c$.

$y = \left(4 x - 15\right) \left(2 x - 2\right) - {\left(3 x - 1\right)}^{2}$

The first thing we have to do is distribute and expand:
$4 x \cdot 2 x = 8 {x}^{2}$

$4 x \cdot - 2 = - 8 x$

$- 15 \cdot 2 x = - 30 x$

$- 15 \cdot - 2 = 30$

When we combine this all together we get:
$8 {x}^{2} - 8 x - 30 x + 30$

We can still combine like terms by doing $- 8 x - 30 x$:
$8 {x}^{2} - 38 x + 30$

$- - - - - - - - - - - - - - - - - - -$

Now let's look at ${\left(3 x - 1\right)}^{2}$ and expand:
$\left(3 x - 1\right) \left(3 x - 1\right)$

$3 x \cdot 3 x = 9 {x}^{2}$

$3 x \cdot - 1 = - 3 x$

$- 1 \cdot 3 x = - 3 x$

$- 1 \cdot - 1 = 1$

When we combine this all together we get:
$9 {x}^{2} - 3 x - 3 x + 1$

Then we combine like terms by doing $- 3 x - 3 x$:
$9 {x}^{2} - 6 x + 1$

$- - - - - - - - - - - - - - - - - -$

So the equation is now:
$y = 8 {x}^{2} - 38 x + 30 - \left(9 {x}^{2} - 6 x + 1\right)$

Let's distribute the negative sign:
$y = 8 {x}^{2} - 38 x + 30 - 9 {x}^{2} + 6 x - 1$

Finally, let's combine like terms again:
$y = \textcolor{red}{8 {x}^{2}} \quad \textcolor{m a \ge n t a}{- \quad 38 x} + \textcolor{b l u e}{30} \quad \textcolor{red}{- \quad 9 {x}^{2}} + \textcolor{m a \ge n t a}{6 x} \quad \textcolor{b l u e}{- \quad 1}$

So the final answer in standard form is:
$y = - {x}^{2} - 32 x + 29$
as it matches $y = a {x}^{2} + b x + c$.

Hope this helps!