Question

Can somebody help me figure out by factoring? while listing the steps so I can understand how you did it? Thanks!:)
3(5-x)^4+2(5-x)^3-(5-x)^2

Answer 1

Hey :)
\[\large\rm 3(5-x)^4+2(5-x)^3-(5-x)^2\]

Answer 2

\[\large\rm 3(5-x)(5-x)(5-x)^2+2(5-x)(5-x)^2-(5-x)^2\]The first one is a 4th power,
so I can pull a couple of them out
(this isn't the factoring,
I'm just trying to get a feel for what I'll be factoring out of each term.
I'm just applying this idea right now: \(\large\rm x^4=x\cdot x\cdot x^2\))

Answer 3

From here we can clearly see what they have in common:\[\large\rm 3(5-x)(5-x)\color{#DD4747}{(5-x)^2}+2(5-x)\color{#DD4747}{(5-x)^2}-\color{#DD4747}{(5-x)^2}\]

Answer 4

Oh let's add the brackets on the outside of everything,
maybe that will make things a little more clear.\[\large\rm \left[3(5-x)(5-x)\color{#DD4747}{(5-x)^2}+2(5-x)\color{#DD4747}{(5-x)^2}-\color{#DD4747}{(5-x)^2}\right]\]So we're doing is,
we're pulling this common factor out of the square brackets.\[\large\rm =\color{#DD4747}{(5-x)^2}\left[3(5-x)(5-x)+2(5-x)-?\right]\]Gotta be careful here.
What am I going to get in my last term? :)
When you fully factor something out of itself.

Answer 5

why did you keep (5-x)^2 at the end? why not just write it out another time?

Answer 6

is that a GCF?

Answer 7

Yes.
It appears that (5-x)^2 was the GCF!
So we didn't really want to break it down into (5-x)(5-x),
you could though :)

Answer 8

okaayyyy i got it, now would the -? at the end be -1?

Answer 9

Good!
When you divided (5-x)^2 out of (5-x)^2, you're left with a 1.
You're not left with a 0,
I just wanted to make sure that was clear.\[\large\rm =\color{#DD4747}{(5-x)^2}\left[3(5-x)(5-x)+2(5-x)-1\right]\]

Answer 10

okay so now what? can it be simplified further?

Answer 11

\[\large\rm =(5-x)^2\left[\color{royalblue}{3(5-x)(5-x)+2(5-x)-1}\right]\]
`Maybe`.
We'll have to do the work of expanding out all of this blue nonsense before we can determine whether it can be factored further.

Answer 12

Do you understand how to multiply (5-x)(5-x)? :)
Think you can get the blue stuff into the form ax^2+bx+c?

Answer 13

I understand that they are binomials and can be simplified by using FOIL correct? or do we not use this equation for this?

Answer 14

Good :)

We've done the GCF method already, there isn't any more scraps to pull out of there.
Now we have to rely on our FOIL method.

Answer 15

so what will it look like after that is done?

Answer 16

\[\large\rm =(5-x)^2\left[3\color{royalblue}{(5-x)(5-x)}+2(5-x)-1\right]\]FOIL'ing out the (5-x) brackets gives us:\[\large\rm =(5-x)^2\left[3\color{royalblue}{(25-10x+x^2)}+2(5-x)-1\right]\]

Answer 17

Now we have to go a little further,
distributing the 3 to the blue stuff,
distributing the 2 to the next set of brackets.

Answer 18

ok ill try it and compare my answer with yours!:)

Answer 19

ok im ready!

Answer 20

\[\large\rm =(5-x)^2\left[75-30x+3x^2+10-2x-1\right]\]So I guess we get something like this, ya?

Answer 21

After that, we need to combine like-terms.

Answer 22

i got that also!

Answer 23

yay!
now try to combine like-terms :D

Answer 24

I got (5-x)^2[84-32x+3x^2]

Answer 25

Ok good.
Let's write it in standard form, with the x^2 coming first,
\[\large\rm =(5-x)^2\left[3x^2-32x+84\right]\]

Answer 26

Hmm this step is going to be a bit tricky.
Do you remember `factor by grouping`?
That's what we need to apply here.

Answer 27

no I don't... can you show me what that would look like?

Answer 28

When we're given a standard second degree polynomial,\[\large\rm ax^2+bx+c\]We multiply \(\large\rm a\) and \(\large\rm c\) and try to find factors of \(\large\rm ac\) that add to \(\large\rm b\).

Answer 29

So for this problem...
We're multiplying \(\large\rm 3\) and \(\large\rm 84\)
and trying to find two factors of that number that add to \(\large\rm -32\)

Answer 30

We don't want to deal with a big number like 3*84,
so instead we'll try a neat trick.
Let's break 84 into its prime factorization.

Answer 31

\[\large\rm 3\cdot\color{orangered}{84}=3\cdot\color{orangered}{2\cdot42}=3\cdot\color{orangered}{2\cdot7\cdot6}\]Do you understand how I broke that down? :O

Answer 32

yes, now how did you know to stop breaking it down at 6?

Answer 33

Ooo good question.
I could have gone further, I didn't actually succeed in getting all the primes lol.

All I was really trying to do was break it into "small" numbers,
so we can `test` some combinations.

Answer 34

WAIT

Answer 35

But yes, let's break the 6 up, just in case it's necessary.\[\large\rm =2\cdot2\cdot3\cdot3\cdot7\]

Answer 36

is it (3x-28)(x-3) ?

Answer 37

Sec, checking :)

Answer 38

Hmmm no :d

Answer 39

why not? where did i go wrong?

Answer 40

If you FOIL your result back out:
You get middle terms of: \(\large\rm -28x\) and \(\large\rm -9x\)
which unfortunately don't get us to \(\large\rm -32x\)
:(

Answer 41

wow a simple subraction mistake lol my b!

Answer 42

i got ahead of myself...

Answer 43

hehe

Answer 44

Let's try some combinations.
How bout \(\large\rm 2\cdot2\cdot3\) and \(\large\rm 3\cdot7\)

Answer 45

So that gives us \(\large\rm 12\) and \(\large\rm 21\)
Hmm, no way to get \(\large\rm 32\) when combining those.
So we need to try another combination.

Answer 46

Maybe this combination instead?
\(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\)

Answer 47

no that didn't work, but would the combination of 6 and 14 work? with 14 being in the first group? (3x-14)(x-6)

Answer 48

Woops! :)
Hold on, let's look at my last combination a sec.
I think it works.

Answer 49

Given \(\large\rm 2\cdot3\cdot3\) and \(\large\rm 2\cdot7\),

So we have \(\large\rm 18\) and \(\large\rm 14\) ya?
Any way to combine these and get \(\large\rm 32\)?

Answer 50

Oh you got to the right result,
you must have some sneaky shortcut method :P
I'm not sure what you did lol.

Answer 51

no lol just trial and erro method:P

Answer 52

but now what would it look like if i wrote it out completely?

Answer 53

\(\large\rm 18+14=32\)
Which tells us that 18 and 14 are the factors that we want!
So we break up the -32x into -18x and -14x and do factor by grouping.\[\large\rm 3x^2\color{orangered}{-32x}+84\]\[\large\rm 3x^2\color{orangered}{-18x-14x}+84\]Factor a 3x out of each of the first two terms,\[\large\rm 3x(x-6)-14x+84\]Pull a -14 out of each of the other terms,\[\large\rm 3x(x-6)-14(x-6)\]Then pull an (x-6) out of everything,\[\large\rm (x-6)(3x-14)\]

Answer 54

I know we're kind of past that step already,
but I just wanted to detail the `factor by grouping` method in case you needed to see it.

Answer 55

thank you! thats this is the explanation i was hoping for!

Answer 56

So we've gotten to this point:\[\large\rm =(5-x)^2\left[\color{orangered}{3x^2-32x+84}\right]\]\[\large\rm =(5-x)^2\left[\color{orangered}{(3x-14)(x-6)}\right]\]We can drop the square brackets at this point as they're searching no purpose.\[\huge\rm (5-x)^2(3x-14)(x-6)\]And we're finally done!
yayyy team! \c:/

Answer 57

serving# no purpose

Answer 58

aye!!!! thank you so much!

Answer 59

typo :p

Answer 60

np :3