# What is the standard form of  y= (x+3)(x+1)+(3x-7)^2?

Apr 18, 2016

Standard form is $y = 10 {x}^{2} - 38 x + 52$

#### Explanation:

As this a quadratic equation, standard form of this is

$y = a {x}^{2} + b x + c$

Hence simplifying $y = \left(x + 3\right) \left(x + 1\right) + {\left(3 x - 7\right)}^{2}$

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

= $\left({x}^{2} + 3 x + x + 3\right) + \left(9 {x}^{2} - 42 x + 49\right)$

= $10 {x}^{2} - 38 x + 52$