Question

FACTORING BY GROUP:
1. y^2 - 8y - y + 8

2. 25 - (x^2-6x+9)

3. x^2 + 4xy + 4y^2 - 64

Answer 1

1) Factorize the first two terms and last two terms, and try to factorize even further.
2) First, distribute the negative sign. Then, find such an arrangement that you are able to factorize the first and last two terms with the same in-bracket thing. Now see 1.
3) First two and last two :D

Answer 2

1)y^2 - 8y - y + 8=y^2-9y+8=(y-1)(y-8)

Answer 3

\[ \frac{ y ^{2} }{ y ^{} }-\frac{ 8y }{ y }-(y-8)\]

Answer 4

OK I got it

Answer 5

Let me try the others

Answer 6

2. 25 - (x^2-6x+9)=25-x^2+6x-9=-x^2+6x+16=-(x^2-6x-16)=-(x-8)(x+2)=(8-x)(x+2)

Answer 7

3. x^2 + 4xy + 4y^2 - 64=(x+2y)^2-64=(x+2y)^2-8^2=(x+2y+8)(x+2y-8)
hope it helps!

Answer 8

I'm confused about problem 2 and 3

Answer 9

which step?

Answer 10

Lets begin with problem 2. I did as Looking for trinomials. SO I got (5-x+3)(5+x-3)

Answer 11

I'm still confused. Looking at examples and trying to figure out. Also #3. I don't get it

Answer 12

(5-x+3)(5+x-3)
5-x+8-x
5+x-3=2+x
(5-x+3)(5+x-3)=(8-x)(2-x)
You get the same answer as @mymathcourses even if u do in a different way bcause that is the simplified answer.:)
Getting it @rosimary?

Answer 13

Yes. I'm understanding

Answer 14

I was confused because the sample the teacher gave for that particular problem she stopped at (4-x+y)(4+x-y). I guess she is teaching us by step? I'm not sure but I'm understanding.

Answer 15

I am still confused as to what to do about problem 3. I got a weird answer that looks like (x+1)(x+4y)(2y-8)(2y+8)

Answer 16

yes , it should use a^2- b^2=(a+b)(a-b)

Answer 17

a^2 year is (x+Y)^2 and b^2 here is 8^2

Answer 18

which problem are you ref too?