What is the standard form of #y= (x+6)(x-3)(x+2)#?

1 Answer
Apr 23, 2017

See the entire solution process below:

Explanation:

First, multiply the two right most terms within parenthesis. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#y = (x + 6)(color(red)(x) - color(red)(3))(color(blue)(x) + color(blue)(2))# becomes:

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

We can now combine like terms:

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

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

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

Now, we again multiply the two terms in parenthesis on the right side of the equation:

#y = (color(red)(x) + color(red)(6))(color(blue)(x^2) - color(blue)(1x) - color(blue)(6))# becomes:

#y = (color(red)(x) xx color(blue)(x^2)) - (color(red)(x) xx color(blue)(1x)) - (color(red)(x) xx color(blue)(6)) + (color(red)(6) xx color(blue)(x^2)) - (color(red)(6) xx color(blue)(1x)) - (color(red)(6) xx color(blue)(6))#

#y = x^3 - 1x^2 - 6x + 6x^2 - 6x - 36#

We can group and combine like terms to put the equation into standard form:

#y = x^3 + 6x^2 - 1x^2 - 6x - 6x - 36#

#y = x^3 + (6 - 1)x^2 + (-6 - 6)x - 36#

#y = x^3 + 5x^2 + (-12)x - 36#

#y = x^3 + 5x^2 - 12x - 36#