How do you find the product of #(2h+3)(2h^2+3h+4)#?

2 Answers
Jim G.
Sep 7, 2016

#4h^3+12h^2+17h+12#

Explanation:

Each term in the 2nd bracket must be multiplied by each term in the 1st bracket.

The following shows how this may be done.

#(color(red)(2h+3))(2h^2+3h+4)#

#=color(red)(2h)(2h^2+3h+4)color(red)(+3)(2h^2+3h+4)#

now distribute each pair of brackets.

#=4h^3+6h^2+8h+6h^2+9h+12#

and collecting like terms gives.

#=4h^3+(6h^2+6h^2)+(8h+9h)+12#

#=4h^3+12h^2+17h+12larr"result of product"#

Alan P.
Sep 7, 2016

#(2h+3)(2h^2+3h+4)=color(orange)(4h^3+12h^2+17h+12)#

Explanation:

#color(red)(""(2h+3))(2h^2+3h+4)#

using the distributive property
#color(white)("XXX")=color(red)(2h)(2h^2+3h+4)color(red)(+3)(2h^2+3h+4)#

(no real change except color to see the separate terms more clearly)
#color(white)("XXX")=color(blue)(2h(2h^2+3h+4))+color(green)(3(2h^2+3h+4)#

expanding each term
#color(white)("XXX")=color(blue)(4h^3+6h^2+8h)+color(green)(6h^2+9h+12)#

grouping equal powers of the variable
#color(white)("XXX")=color(blue)(4h^3)+(color(blue)(6h^2)+color(green)(6h^2))+(color(blue)(8h)+color(green)(9h))+(color(green)(12))#

Adding together equal powers of the variable
#color(white)("XXX")=4h^3+12h^2+17h+12#

Related questions
See all questions in Multiplication of Polynomials by Binomials
Impact of this question
4283 views around the world
You can reuse this answer
Creative Commons License