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

##### 1 Answer

$- {x}^{2} + 10 x - 14$

#### Explanation:

Let's first do the multiplication of the brackets using FOIL, then add in the remaining terms:

FOIL

• $\textcolor{red}{F}$ - First terms - $\left(\textcolor{red}{a} + b\right) \left(\textcolor{red}{c} + d\right)$
• $\textcolor{b r o w n}{O}$ - Outside terms - $\left(\textcolor{b r o w n}{a} + b\right) \left(c + \textcolor{b r o w n}{d}\right)$
• $\textcolor{b l u e}{I}$ - Inside terms - $\left(a + \textcolor{b l u e}{b}\right) \left(\textcolor{b l u e}{c} + d\right)$
• $\textcolor{g r e e n}{L}$ - Last terms - $\left(a + \textcolor{g r e e n}{b}\right) \left(c + \textcolor{g r e e n}{d}\right)$

and so $\left(x - 3\right) \left(4 - x\right)$ becomes:

• $\textcolor{red}{F} = 4 x$
• $\textcolor{b r o w n}{O} = - {x}^{2}$
• $\textcolor{b l u e}{I} = - 12$
• $\textcolor{g r e e n}{L} = 3 x$

which adds to:

$4 x - {x}^{2} - 12 + 3 x = - {x}^{2} + 7 x - 12$

Now let's add in the remaining terms:

$\left(- {x}^{2} + 7 x - 12\right) + 3 x - 2 = - {x}^{2} + 10 x - 14$