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

1 Answer
Douglas K.
Apr 26, 2017

Please see the explanation.

Explanation:

The method for performing the multiplication will become obvious when we write it in a form where each term of the first factor multiplies the second factor:

#(2t^2+t+3)(4t^2+2t-2) = #

#2t^2(4t^2+2t-2) +#

#t(4t^2+2t-2) + #

#3(4t^2+2t-2)#

Distribute the first term of the first factor into the second factor:

#(2t^2+t+3)(4t^2+2t-2) = #

#8t^4+4t^3-4t^2 +#

#t(4t^2+2t-2) + #

#3(4t^2+2t-2)#

Distribute the second term of the first factor into the second factor:

#(2t^2+t+3)(4t^2+2t-2) = #

#8t^4+4t^3-4t^2 +#

#color(white)(".........")4t^3+2t^2-2t + #

#3(4t^2+2t-2)#

Distribute the third term of the first factor into the second factor:

#(2t^2+t+3)(4t^2+2t-2) = #

#8t^4+4t^3-4t^2 +#

#color(white)(".........")4t^3+2t^2-2t + #

#color(white)(".................")12t^2+6t-6#

I have aligned the columns so that they can be added:

#(2t^2+t+3)(4t^2+2t-2) = #

#8t^4+4t^3-4t^2 +#

#color(white)(".........")4t^3+2t^2-2t + #

#ul(color(white)(".................")12t^2+6t-6)#
#8t^4+8t^3+10t^2+4t-6#

Because multiplication is commutative, we could have multiplied the first factor by each term of the second factor and obtained the same result.

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