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

Question

Avery Alexander · Accepted Answer

See the entire solution process below:

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#

Kennedy Adams · Answer

undefined To write the equation $ y = 4(x+1)(x+2) $, you need to apply the distributive property to expand the expression. First, distribute the $4$ to both terms inside the parentheses:

$ y = 4(x^2 + 2x + x + 2) $

Next, simplify by combining like terms:

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

So, the expanded form of the given expression is $ y = 4x^2 + 12x + 8 $.