How do you multiply #(2y - 7) ( 3y ^ { 2} + 6y - 5)#?

1 Answer
George C.
Nov 9, 2016

#(2y-7)(3y^2+6y-5) = 6y^3+9y^2-52y+35#

Explanation:

The formal way of doing the multiplication goes like this:

#(2y-7)(3y^2+6y-5) = 2y(3y^2+6y-5)-7(3y^2+6y-5)#

#color(white)((2y-7)(3y^2+6y-5)) = (6y^3+12y^2-10y)-(21y^2+42y-35)#

#color(white)((2y-7)(3y^2+6y-5)) = 6y^3+12y^2-21y^2-10y-42y+35#

#color(white)((2y-7)(3y^2+6y-5)) = 6y^3+9y^2-52y+35#

Personally, I would just look at each possible power of #y# from #y^3# down to #y^0# in turn and total up the products of pairs of terms that contribute to it:

#y^3:" " 2*3 = 6#

#y^2:" " 2*6-7*3 = 12-21 = -9#

#y:" " 2(-5)-7*6 = -10-42 = -52#

#1:" " (-7)(-5) = 35#

Hence:

#6y^3-9y^2-52y+35#

With practice, it's usually not too hard to do each coefficient in your head and just write them down one at a time.

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