What is the standard form of  y= (x-7)(3x-5)-(x+7)^2?

Nov 16, 2015

So the standard form is $a {x}^{2} + b x + c$

Explanation:

$\textcolor{red}{\underline{\text{You asked for 'standard' form}}}$

Consider the first part:
$\left(x - 7\right) \left(3 x - 5\right) \to 3 {x}^{2} - 5 x - 21 x + 35 = 3 {x}^{2} - 26 x + 35$

Consider the second part:
${\left(x - 7\right)}^{2} \to {x}^{2} - 14 x + 49$

Put them together and we end up with:

$y = \left(3 {x}^{2} - 26 x + 35\right) - \left({x}^{2} - 14 x + 49\right)$

The minus sign outside the brackets reverses all the signs within.

$y = 3 {x}^{2} - 26 x + 35 - {x}^{2} + 14 x - 49$

$\textcolor{b r o w n}{y = 2 {x}^{2} - 12 x - 14}$