Question

@hero Factoring

Answer 1

The expression \[x^2-5xy-176y^2\] is equivalent to
(x-16y)(x+11y)
(x-8y)(x-22y)
(x-8y)(x+22y)
(x+8y)(x-22y)
(x+8y)(x+22y)

Answer 2

If you are given a trinomial of the form \(ax^2 + bx + c\) and asked to factor it into some form of \((cx + d)(fx + g)\), you must first find two numbers that add to get \(b\) yet multply to get \(ac\)

For example, if you are given something like \(3x^2 + 5x + 2\) do you know how to factor that @EndersWorld

Answer 3

In factor boxes... I can’t do it in my head.

Answer 4

Can you do it using any method you've been shown so far?

Answer 5

@EndersWorld, take your time

Answer 6

No. Boxes don’t work out.

Answer 7

Ah, okay, good. Now that I have your complete undivided attention and interrupting forces are gone...

Answer 8

^_^

Answer 9

Here's how to factor 3x^2 + 5x + 2 ....

@jhonyy9 was on the right track, just not explaining completely enough. You do want to find two numbers that add to 5, but you also want to find two numbers that multiply to get (3)(2) or 6. Keep in mind, b = 5, a = 3, c = 2.

So in general, you want to find two numbers that add to get b, yet multiply to get ac.

You can set it up this way:
If m and n represent the two numbers you wish to find then:
m + n = 5
m x n = 6

Answer 10

Correct.

Answer 11

Beginning with factors of 6:
(6)(1)
(3)(2)

So those are the only two unique factors. Now you think to yourself, which of these would ADD to get 5. Only 3 and 2 would work in this case. so m = 3 and n = 2. Now you take that and REPLACE the b value (5) with (3 + 2) like so

3x^2 + 5x + 2 becomes 3x^2 + (3 + 2)x + 2

Next distribute the x:
3x^2 + 3x + 2x + 2

Afterwards, "Factor by Grouping" which means factor the first two terms then factor the last two terms like so:
3x is common to the first to terms and 2 is common to the last two terms therefore you should get 3x(x + 1) + 2(x + 1) as the next step in the factorization process. With me so far?

Answer 12

I think!

Answer 13

Okay, I'll take that as a yes.

so in the previous step we have 3x(x + 1) + 2(x + 1) and we still need to factor further to get the form (cx + d)(fx + g) so what do we do? Well remember what I was explaining to you earlier about the distributive property? Well guess what? In this case our a term is
x + 1, our b term is 3x and our c term is 2. Remembering the distributive property a(b + c) = ac + ac, we can use that in reverse to complete the factorization. If we do that, we end up with (x + 1)(3x + 2) as the final factorization. Open to any questions you have about this.

Answer 14

Why does this exist?

Answer 15

Why does WHAT exist? Factoring? Its one of the fundamental principles of mathematics a prerequisite to higher level mathematics such as calculus, probablity theory, statistics, calculus, linear algebra, discrete mathematics, abstract algebra, differential equations, topology, etc.

Answer 16

Anyway, to help you understand more simply, I'll post the complete steps without all the explanation part so that you see the process more completely:

Answer 17

I think that would be beneficial

Answer 18

3x^2 + 5x + 2 = 3x^2 + (3 + 2)x + 2
= 3x^2 + 3x + 2x + 2
= 3x(x + 1) + 2(x + 1)
= (x + 1)(3x + 2) <-- Application of distributive property (in reverse)

Answer 19

Nice handwriting ( ͡° ͜ʖ ͡°) :)

Answer 20

Even further explanation here for each step:

Original Problem:
3x^2 + 5x + 2

Replace 5 with (3 + 2) since m = 3 and n = 2:
3x^2 + (3 + 2)x + 2

Distribute the x using distributive property:
3x^2 + 3x + 2x + 2

Factor by Grouping (3x is common to the 1st 2 terms, 2 is common to the last 2):
3x(x + 1) + 2(x + 1)

And remember the "a" term is (x+1) which is common to both factorizations so factor that out as well to get:
= (x + 1)(3x + 2)

Answer 21

Ima try the problem now.

Answer 22

Good luck although I must say, it is a slightly different problem.

Answer 23

You better stick around XD

Answer 24

Since b is -5 should I just do like -2+-3?

Answer 25

Well, this one is definitely a slightly different problem and for it, I would use a different method. It's a bit more complicated than what I have already shown you thus far. I don't know if you'd be able to keep up.

Answer 26

Should I try using factor boxes?

Answer 27

Nevertheless, you don't have to worry about the a term since it equals 1 in this case, so you should be able to use factors of -176 that add to get -5.

Answer 28

-16 and 11

Answer 29

By the way, if they did not give you the answer choices and you had to find the factors of 176, here's what I would do if I were you. Factor like so, ALWAYS start with 1:

(1)(176)
(2)(88)

is 88 divisible by 3, no.
is 88 divisible by 4, yes:
4(44)

44 is divisible by 4 so
(4)(4)(11)

Now we only need two factorizations so:
(16)(11)

But what we have needs to add to -5 so how can 16 and 11 add to get -5. Well, as you put it above 16 must be negative so the two factors are -16 and 11.

Answer 30

You should be able to factor from there.

Answer 31

(x-16y)(x+11y)

Answer 32

I don’t have your answer as a answer choice :0

Answer 33

Sorry, I forgot the y, just a typo (x - 11y)(x + 16y)

Answer 34

Oh, I messed up further than that I think. Hang on.

Answer 35

Oh I see where I messed up. Sorry about that.

Answer 36

The 5 in the original problem, the factors of that are 16 and -11 which is what I replaced in the original problem so what I should have done was this:

Answer 37

x^2 - 5xy - 176y^2 = x^2 - (16 - 11)xy - 176y^2
= x^2 -(16xy - 11xy) - 176y^2
= x^2 - 16xy + 11xy - 176y^2
= x(x - 16y) + 11y(x - 16y)
= (x - 16y)(x + 11y)

Answer 38

There we go. Good catch. I was typing too fast and didn't catch myself making that mistake to begin with. When I factor these I usually don't mess with the negative number until I absolutely have to.

Answer 39

Same. I hate it.

Answer 40

Yeah, but I don't usually make that mistake. I just haven't done these in a while. You reminded me of my own fail safe. Never deal with the negative number until you absolutely have to. If you use my method correctly, you can just find the positive factors of 176 and then replace that with the positive 5 but leave the negative on the outside of the parentheses and distribute it in later.

Answer 41

Moving onto the next one :0

Answer 42

Yep, that was probably a good stopping point.