# What is the standard form of  y= (2x+14)(x+12)-(7x-7)^2?

Aug 30, 2017

$y = - 47 {x}^{2} + 136 x + 119$

#### Explanation:

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

$y = 2 {x}^{2} + 24 x + 14 x + 168 - \left(49 {x}^{2} - 98 x + 49\right)$

$y = 2 {x}^{2} + 24 x + 14 x + 168 - 49 {x}^{2} + 98 x - 49$

$y = - 47 {x}^{2} + 136 x + 119$

Aug 30, 2017

$y = - 47 {x}^{2} + 136 x + 119$

#### Explanation:

The equation of a quadratic in standard form is: $y = a {x}^{2} + b x + c$

So, this question is asking us to find $a , b , c$

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

It is probably simplier to break $y$ in its two parts first.

$y = {y}_{1} - {y}_{2}$

Where: ${y}_{1} = \left(2 x + 14\right) \left(x + 12\right)$ and ${y}_{2} = {\left(7 x - 7\right)}^{2}$

Now, expand ${y}_{1}$

${y}_{1} = 2 {x}^{2} + 24 x + 14 x + 168$

$= 2 {x}^{2} + 38 x + 168$

Now, expand ${y}_{2}$

${y}_{2} = {\left(7 x - 7\right)}^{2} = {7}^{2} {\left(x - 1\right)}^{2}$

$= 49 \left({x}^{2} - 2 x + 1\right)$

$= 49 {x}^{2} - 98 x + 49$

We can now simply combine ${y}_{1} - {y}_{2}$ to form $y$

Thus, $y = 2 {x}^{2} + 38 x + 168 - \left(49 {x}^{2} - 98 x + 49\right)$

$= 2 {x}^{2} + 38 x + 168 - 49 {x}^{2} + 98 x - 49$

Combine coefficients of like terms.

$y = \left(2 - 49\right) {x}^{2} + \left(38 + 98\right) x + \left(168 - 49\right)$

$y = - 47 {x}^{2} + 136 x + 119$ (Is our quadratic in standard form)

$a = - 47 , b = + 136 , c = + 119$