How do you write # y=-3(x-3)(x+2)# in standard form?

1 Answer
Jul 25, 2017

See a solution process below:

Explanation:

First, we need to multiply the two terms with parenthesis. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#y = -3(color(red)(x) - color(red)(3))(color(blue)(x) + color(blue)(2))# becomes:

#y = -3((color(red)(x) xx color(blue)(x)) + (color(red)(x) xx color(blue)(2)) - (color(red)(3) xx color(blue)(x)) - (color(red)(3) xx color(blue)(2)))#

#y = -3(x^2 + 2x - 3x - 6)#

We can next combine like terms within the parenthesis:

#y = -3(x^2 + [2 - 3]x - 6)#

#y = -3(x^2 + [-1]x - 6)#

#y = -3(x^2 - 1x - 6)#

#y = -3(x^2 - x - 6)#

Now, we can multiply each term within the parenthesis by the term outside the parenthesis:

#y = color(red)(-3)(x^2 - x - 6)#

#y = (color(red)(-3) * x^2) - (color(red)(-3) *x) - (color(red)(-3) *6)#

#y = -3x^2 - (-3x) - (-18)#

#y = -3x^2 + 3x + 18#