Question

This is algebra and can someone please help me on how to do this? Clueless. okay...
27(a^6)(b^4)+12(a^2)(B^3)

Answer 1

\(27(a^6)(b^4)+12(a^2)(b^3)
\\3a^2b^3(9a^4b+4)
\)

Answer 2

but how though? I mean I trust your answer it's just they did a terrible job of teaching me?

Answer 3

alright, let me think about how to explain this.
so when you divide exponents with the same base it works like this: \(\dfrac{a^M}{a^N}=a^{(M-N)}\)

and so when we factor, we're dividing by the greatest common factor.

for example the factors of
27: 1 3 9 27
12 1 2 3 4 6 12
so we divide each term by 3 and we're left with

\(27(a^6)(b^4)+12(a^2)(b^3) \iff \dfrac{27}{3}(a^6)(b^4)+\dfrac{12}{3} (a^2)(b^3)
\\3(9a^6b^4+4a^2b^2)
\)

Answer 4

factor this: \(2y+8y^3\)

Answer 5

2y(6+y) ??

Answer 6

close. \(2y(1+4y^2)\)
I'll step through it now

Answer 7

How...

Answer 8

let's start with just the numbers 2 and 8,
the factors of 2:
1 2
1 2 4 8

so what is the greatest common factor?

Answer 9

2

Answer 10

so now we divide each term by 2

Answer 11

my bad i was getting a snack.

Answer 12

share :(

Answer 13

\(2y+8y^3\implies \dfrac{2y}{2}+\dfrac{8y^3}{2}\\2(y+4y^3)\)

there's another step we can do here

Answer 14

~gives half of my chicken cordon blu~

Answer 15

I do not under stand how to do that..

Answer 16

what does \(y^2\) mean?

Answer 17

y*y

Answer 18

and so we have 2 terms here
a) y
b) y*y

we have a common term of y that we can pull out

Answer 19

you jusy y....

Answer 20

huh

Answer 21

I Am 50%*2 confused...

Answer 22

\(2y(\dfrac{y}{y}+\dfrac{4y^3}{y})\)
in order to pull out a y, we have to divide each term by y. and so now we simplify the stuff int he parentheses and we get

\(2y(1+4y)

Answer 23

\(2y(1+4y)\)

Answer 24

So, y/y= 1 and if we cancel out the ys on the other we get... 4 y...
so it would be 2y(1+4y)

Answer 25

and so that's the basic process we do for your problem.