+0

# Factoring

+1
244
3
+2448

Sep 5, 2018

#1
+67
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)

Sep 5, 2018
#2
+2448
+1

I need to know how to get each of them

RainbowPanda  Sep 5, 2018
#3
+101745
+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

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

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

.
Sep 6, 2018
edited by Melody  Sep 6, 2018