Question

Can someone help me out with this. I dont want the answer. i just want help figuring it out. I'll post it below.

Answer 1

\[\frac{ 2a^2+a-6 }{ a^2+4a-5 }\times \frac{ a^3-3a^2+2a }{ 4a^2-6a }\]

Answer 2

This is an exercise in factoring the polynomials.

can you factor each one? (or any of them!?)

Answer 3

Yes

Answer 4

the "easy" way to multiply is multiply top*top and bottom * bottom
it's clear we will get very ugly polynomial divided by very ugly polynomial

however, if these factor (and they probably do), the same factor in the bottom and the top will cancel. After we cancel all that we can, we multiply out what is left ... probably something relatively simple

Answer 5

thats the problem where i'm at is factoring them.

Answer 6

which ones can you factor?

Answer 7

i got one of bottom ones

Answer 8

can you post what you got?

Answer 9

its 2a(2a-3)

Answer 10

ok. now let's try the other bottom polynomial
a^2 +4a - 5

1) look at the last number and list all pairs that multiply to give you 5
1,5
is the only pair
2) note the sign on the -5. that means the "roots' will have *different* signs
3) look at the +4 on 4a
the + sign means the biggest root is positive
because we have different signs, the smaller root is negative
4) put in the proper signs on the pairs listed in step 1)
-1+5
5) add each pair (in this case we only have -1+5)
+4
6) if the sum matches the +4 in 4a then we found our factors


in this case -1 and 5
write (x-1)(x+5)

Answer 11

let's try
a^3 -3a^2+2a

first, factor a from each term.
what do you get ?

Answer 12

im not very sure with the a^3 and -3^2

Answer 13

remember the little 3 in the upper right is a short cut to show
a*a*a
so we could write it as
a*a*a - 3*a*a +2*a

can you factor a from each term?

Answer 14

yes

Answer 15

what do you get?

Answer 16

I'm not entirely sure.

Answer 17

example of factoring:
if you have
a*a + a
you can divide each term by a: a*a/a = a
and a/a = 1
to get (a+1) multiplied by the a we "factored out"
a(a+1)
if we distribute the a (multiply a times each term inside the parens) we get
a^2 +a

showing that a(a+1) is the same as a^2 +a

Answer 18

a*a*a - 3*a*a +2*a

can you factor a from each term?

Answer 19

we "take" an "a" out of each term (technically we divide) and write the a outside the parens

Answer 20

would it be a or a^2 outside the ( )

Answer 21

you can factor out something that is *shared* by all the terms

if you have
a^3 +a^2

which means
a*a*a + a*a

both terms have a*a , so we can factor out a*a

Answer 22

but notice in
a^3 -3a^2+2a
which means
a*a*a -3*a*a + 2*a

the last term 2*a has only 1 a. Each term has 1 a, but not 2 a's
so we take out an "a" from each term
(not a*a, just a)

Answer 23

would there be two ( ) or one?

Answer 24

just one a can be factored out.

Answer 25

would it be a(a^2-3a+2)

Answer 26

yes

Answer 27

now we can do more. Factor
a^2-3a+2

Answer 28

\[\frac{ (2a-3)(a+2) }{ (a-1)(a+5) }\times \frac{ a(a^2-3a+2) }{ 2a(2a-3) }\] is this what i should have right now

Answer 29

yes that looks good
but you can factor
a^2 -3a + 2

can you use the steps I posted up above?