Simplify #(2x^2+3x+4)(5x^2+7x+6)#?

2 Answers
May 8, 2016

#(2x^2+3x+4)(5x^2+7x+6) = 10x^4+29x^3+53x^2+46x+24#

Explanation:

Since the coefficients are all positive, then if you have a calculator to hand we can calculate the coefficients of the product like this:

#20304 xx 50706 = 1029534624#

How does that help?

Each pair of digits represents a coefficient, so:

#color(purple)(2)color(blue)(03)color(green)(04) xx color(purple)(5)color(blue)(07)color(green)(06) = color(red)(10)color(magenta)(29)color(purple)(53)color(blue)(46)color(green)(24)#

represents:

#(color(purple)(2)x^2+color(blue)(3)x+color(green)(4))(color(purple)(5)x^2+color(blue)(7)x+color(green)(6)) = color(red)(10)x^4+color(magenta)(29)x^3+color(purple)(53)x^2+color(blue)(46)x+color(green)(24)#

This is essentially putting #x=100# in each of the expressions.

Since the largest result we can get by multiplying a one digit number by a one digit number is less than #100# - that is #9xx9 = 81 < 100# - there will be no carries to mess up the pairs of digits in the answer.

May 8, 2016

#(2x^2+3x+4)(5x^2+7x+6)=10x^4+29x^3+53x^2+46x+24#

Explanation:

The formal way of doing the multiplication is something like this:

#(2x^2+3x+4)(5x^2+7x+6)#

#=2x^2(5x^2+7x+6)+3x(5x^2+7x+6)+4(5x^2+7x+6)#

#=((2*5)x^4+(2*7)x^3+(2*6)x^2)+((3*5)x^3+(3*7)x^2+(3*6)x)+((4*5)x^2+(4*7)x+(4*6))#

#=(10x^4+14x^3+12x^2)+(15x^3+21x^2+18x)+(20x^2+28x+24)#

#=10x^4+(14+15)x^3+(12+21+20)x^2+(18+28)x+24#

#=10x^4+29x^3+53x^2+46x+24#

A more compact way of finding the coefficients uses a grid to help us find the combinations of terms to add:

#underline(color(white)(000)|color(white)(00)5color(white)(00)7color(white)(00)6color(white)(00))#
#color(white)(000)|color(white)(0)#
#color(white)(0)2color(white)(0)|color(white)(0)color(red)10color(white)(0)color(magenta)(14)color(white)(0)color(purple)(12)#
#color(white)(000)|color(white)(0)#
#color(white)(0)3color(white)(0)|color(white)(0)color(magenta)(15)color(white)(0)color(purple)(21)color(white)(0)color(blue)(18)#
#color(white)(000)|color(white)(0)#
#color(white)(0)4color(white)(0)|color(white)(0)color(purple)(20)color(white)(0)color(blue)(28)color(white)(0)color(green)(24)#

Each of the numbers in the body of the grid is the product of its row label and column label, e.g. #2xx7=color(magenta)(14)#

Then add the reverse diagonals to find:

#(2x^2+3x+4)(5x^2+7x+6)#

#=color(red)(10)x^4+color(magenta)((14+15))x^3+color(purple)((12+21+20))x^2+color(blue)((18+28))x+color(green)(24)#

#=color(red)(10)x^4+color(magenta)(29)x^3+color(purple)(53)x^2+color(blue)(46)x+color(green)(24)#