Question

Okay so for this one 2a^2-21-65. what do you do with the 2?

Answer 1

It would be in the form (2a )(a ) If you can find factors that work with it. Otherwise, do the quadratic equation.

Answer 2

it would be (2a+13)(a-5). making it 2a=-13 and a=-5. right?

Answer 3

a=5*

Answer 4

Look for a factor pair of ac that sums to b.

That is, ac=2(-65)=-130

Now think about factor pairs of ac that have a sum of b=-21, and then you can factor by grouping.

Answer 5

@ineedhelp111 , your answer is not quite right. When you expand it you get 2x^2+3x-65

Answer 6

Check that factoring with multiplication!! But I think you have the right factor pair.

5*-26=-130
-26+5=-21

Answer 7

so rewrite as \(2x^2-26x+5x-65\)

then factor by grouping. :)

Answer 8

is it a=26 and a=-5?

Answer 9

What did you get when you finished correcting the factoring?

Answer 10

(.....?.....)(.....?.....)

Answer 11

well I remembered the method my teacher showed me how to do with an x and did that and got 26 and 5 I just don't remember whether you flip the symbols or keep them

Answer 12

OK, I'm not sure what you're talking about. We were helping you factor the polynomial. So you should have a product:

(---?---)(---?---)

Answer 13

What you said above
(2a+13)(a-5).

is close but not right. If you FOIL it out, you don't get the polynomial you started with.

Answer 14

so just is the answer a=-26 and a=5 or a=26 and a=-5 or am I completely wrong?

Answer 15

And it makes no difference what the variable is..... a, x, z, r, t, etc... the method is the same. :)

Answer 16

You are close but that is not the answer. That's why I'm trying to find out how you arrived at it, then I can explain what you did wrong.

Answer 17

The last I saw, you said the factoring gave you:

(2a+13)(a-5)

We both told you that wasn't quite right. And even that does not give you either of the answers that you posted. So if you tell me how you got from that factoring, to the answers you posted, I can tell you where the problem is.

Answer 18

I'm assuming that the first thing you did was to fix the factoring....? If so, what did you get for the factored form of the polynomial?

Answer 19

You do this thing where you draw a giant x and in the top part you put in what the first factor and the last factor give you when multiplied and then on the bottom you put the middle factor. then you have to figure out what numbers multiply to the top and add to the bottom one. doing that with this gets you -26 and 5.

Answer 20

(If you do the factoring by grouping like I showed above, it will work out every time. :)

Answer 21

then on the sides is where you write the factors

Answer 22

OK, that's fine - that's similar to the ac method I explained above (I just don't know how the big X fits in, but that's ok - that's just a set-up that someone came up with but the concept is the same).

Now, you have the factor pair -26 and +5 - that's perfect. But now what do you DO with them??

See, if this polynomial had a=1 (no coefficient in front of the x^2) then the factoring is easier, because those number ARE the numbers that go into the (---?---).

But it's different when you have a coefficient in front. That's where factoring by grouping comes in:

Answer 23

how do you factor by grouping?

Answer 24

So like I said above, rewrite as \(2x^2-26x+5x-65\) and then factor by grouping.

\(\large 2x^2-26x+5x-65=2x(x-13)+2(x-13)\)

Answer 25

Now that RHS might not look like it, but it's REALLY just like

ac+bc

You are just letting the c=x-13, follow me?

Answer 26

And of course, ac+bc=c(a+b)

Still true, even if c is a binomial like x-13!

Answer 27

all I got out of that was letters and numbers equal letters and numbers but this is that and more letters and numbers

Answer 28

Sooooo.....

\(\large 2x^2-26x+5x-65=2x(x-13)+2(x-13)=(x-13)(2x+2)\)
^^^^^^^^^
This x-13 is called a BINOMIAL COMMON FACTOR. Because it's a binomial, and it's also a common factor.

Answer 29

@ineedhelp111 , you might have to read what I wrote twice, or three times, or 8 times. I'm going very slowly, just read it carefully and scroll back up as needed. It is pretty basic algebra, and based on the problems you are doing, I'm sure this has been covered in your class (or a previous one). :)

Answer 30

i somehow got a b every quarter in algebra yet the whole year except for a few parts was jibber jabber

Answer 31

Do you understand the factoring that I did, 2 posts up? ^^^

That's how you factor an expression like this. :) It WILL work, every time! Trial and error is fine too, but this is an algorithm, so you can always use it.

Answer 32

No i have no idea what you did and i feel nowhere closer to the answer than i was when i first read the question

Answer 33

Well, I sorta did the factoring part for you, but I see that I accidentally typed a "2" in the final result where I needed a "5"... that's fixed below. But you should be closer to an answer!

The important thing to understand is that factoring is just the "undoing" of multiplication. So you can ALWAYS check it by doing the multiplication and see if you get what you started with. So your factoring above didn't work - if you FOIL it, you'll see what you don't get back to the original expression.

So you have
\(\large 2x^2-26x+5x-65=2x(x-13)+5(x-13)=(x-13)(2x+5)\)

And you want to find where this is =0, so you have

\(\large (x-13)(2x+5)=0\)

Then just solve each of the factors =0, that is find the x that makes
\(x-13=0\) and the one that makes \(2x+5=0\)

Answer 34

I'm sorry, I really have to go as it's very late here. I hope this has been some help!