What is the standard form of y= (x+7)(2x+15)-(x-7)^2y=(x+7)(2x+15)−(x−7)2?

1 Answer
Timber Lin
Dec 24, 2017

y=x^2+43x+56y=x2+43x+56

Explanation:

standard form is y=ax^2+bx+cy=ax2+bx+c

first multiply/distribute to expand everything:
y=(x+7)(2x+15)-(x-7)^2y=(x+7)(2x+15)−(x−7)2
y=x(2x+15)+7(2x+15)-(x-7)(x-7)y=x(2x+15)+7(2x+15)−(x−7)(x−7)
y=2x^2+15x+14x+105-(x(x-7)-7(x-7))y=2x2+15x+14x+105−(x(x−7)−7(x−7))
y=2x^2+29x+105-(x^2-7x-7x+49)y=2x2+29x+105−(x2−7x−7x+49)

combine like terms as you go

y=2x^2+29x+105-x^2+14x-49y=2x2+29x+105−x2+14x−49
y=x^2+43x+56y=x2+43x+56

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