+0  
 
+1
184
3
avatar+2448 

Please help asap

 Sep 5, 2018
 #1
avatar+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
avatar+2448 
+1

I need to know how to get each of them

RainbowPanda  Sep 5, 2018
 #3
avatar+99384 
+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

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