Question

@InYourHead @inkyvoyd @irishchelsc @iamanocean @some_someone @stabar @SnuggieLad @Will679 @dusty can someone help me i dont want the answers i want to learn

Answer 1

the first question is x^2+20x+36

Answer 2

um i suck at math basically so idk so sry bb </3

Answer 3

factor

Answer 4

note
(x+a)(x+b) - this will be the form of your solution

(x+a)(x+b)=x^2+(a+b)x+ab = x^2+20x+36

Answer 5

so given that,

we can conclude that
a+b=20
a*b= 36

then solve for a and b

Answer 6

u just lost me

Answer 7

Can I step in for a moment?

Let's look at our problem here....

\[x^{2}+20x+36\]

You see that number?
It's 36.

And see that middle number?
It's 20.

So, we have to choose two numbers ....
that MULTIPLY to make 36,
and ADD TOGETHER to make 20.

Can you think of two numbers that would fit both of those rules?

Answer 8

(x+a)(x+b)
when you foil that, multiply it out, etc

you will get x^2+ (a+b) x +a*b

Answer 9

That LAST number is 36.
I'm sorry, I forgot to mention "last."

Answer 10

\[x^2+20x+36\]
\[(x+2) (x+18)\]

Answer 11

6X6????

Answer 12

Good. some_someone figured it out.

The numbers are 18 and 2.

18 times 2 is 36.
18 plus 2 is 20.

You get it?

Answer 13

6 times 6 IS 6.

But 6 + 6 = 12.
Not 20.

Good try though.

Answer 14

sorry i didnt know u had to use it for both ?

Answer 15

Yes.

Answer 16

Once we figure out the two numbers,
we write it out like this:

(x + 18) (x + 2)

Answer 17

okay

Answer 18

This was a very easy one problem.

Do you have anymore like it?
If not, I can give you another example, and we can see if you can solve it on your own.

Answer 19

Sorry about the typo.

Answer 20

try 20 more lol

Answer 21

Damn.

Answer 22

Okay. Give us another.

Answer 23

the next one is factor the polynomial 8x^4+6x^3-2x^2

Answer 24

(reverse) distributive property of multiplication
what common multiple do they all share?

Answer 25

note
they're all even
they all have variable x

Answer 26

24

Answer 27

sorry i meant common factor

Answer 28

@completeidiot

Sorry, I'm having trouble with this one.

Is the common factor \[2x^{2}\] ?

Answer 29

yes, you can factor out a \[2x^2\]
\[8x^4+6x^3-2x^2=2x^2(4x^2+3x-1)\]

Answer 30

what's left is factoring
4x^2+3x-1

Answer 31

I solved it.

Thank you very much.

Answer 32

@danya1

Let's leave the \[2x^{2}\] to the side. Forget about that part, for now.

We've broken it down to this....

\[4^{2}+3x-1\]

Answer 33

We have to factor that part now.

Do you know PEMDAS?

Answer 34

Sorry. FOIL.

Answer 35

You know...

First
Outside,
Inside
Last

You know about that?