How do you write #y=4(x+1)(x+2)#?

1 Answer
Apr 19, 2017

See the entire solution process below:

Explanation:

To put this in standard form first multiply the term 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 = 4(color(red)(x) + color(red)(1))(color(blue)(x) + color(blue)(2))# becomes:

#y=4(color(red)(x) xx color(blue)(x)) + (color(red)(x) xx color(blue)(2)) + (color(red)(1) xx color(blue)(x)) + (color(red)(1) xx color(blue)(2)))#

#y = 4(x^2 + 2x + 1x + 2)#

Next, combine like terms:

#y = 4(x^2 + (2 + 1)x + 2)#

#y = 4(x^2 + 3x + 2)#

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

#y = color(red)(4)(x^2 + 3x + 2)#

#y = (color(red)(4) * x^2) + (color(red)(4) * 3x) + (color(red)(4) * 2)#

#y = 4x^2 + 12x + 8#