#1**0 **

9 = (r-2)(2r+sqrt5)(2r-sqrt5)

10 = (7k+6)(3k^2+8)

11 = (6x+7)(sqrt7x+sqrt6)(sqrt7x-sqrt6)

12 = (p+6)(p+sqrt3)((p-sqrt3)

isthebest123
Sep 5, 2018

#3**+2 **

Given a rough inspection I think they are all the same sort of factorization.

So if I do the first one and you understand then you are capable of doing the other three yourself.

9)

\(4r^3-8r^2-5r+10\\ =4r^3-8r^2\qquad\qquad -5r+10\\ \text{now factorize the pairs}\\ =4r^2(r-2)\qquad\qquad -5(r-2)\\\)

now you can think of (r-2) as just one thing.

So you have \(4r^2\) lots of (r-2) and then you take away \(5\) lots of (r-2)

That will give you \((4r^2-5)\) lots of (r-2)

which is

\((4r^2-5)(r-2)\)

you could take this one step further by treating \(4r^2-5\) as the difference of 2 squares

\(4r^2 = (2r)^2 \) and \(5=(\sqrt5)^2 \) so

\((4r^2-5) = (2r-\sqrt5)(2r+\sqrt5)\)

\((4r^2-5)(r-2)=(2r-\sqrt5)(2r+\sqrt5)(r-2)\)

Melody
Sep 6, 2018