Question

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

Answer 1

`(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.