Please help me with factoring 10m^3n^2-15m^2n^1 medal + fan just don't just give me the answer. Explain IT.

- baby456

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- baby456

@mathstudent55

- baby456

@dajakesta

- mathstudent55

\(\large 10m^3n^2-15m^2n^1\)
You are factoring a two-term polynomial.
Start by factoring a common factor out of both terms.
What is the GCF of both terms?

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## More answers

- baby456

5

- mathstudent55

Correct. For the number parts, 10 and -15, you can factor out 5.
What about for the variable parts?
What do m^3 and m^2 have in common?

- baby456

6

- mathstudent55

Not a number. It's a variable to a power.

- mathstudent55

|dw:1435094377519:dw|

- mathstudent55

m^3 and m^2 have m^2 in common.

- mathstudent55

Then we see what n^2 and n^1 have in common.
|dw:1435094442799:dw|
n^2 and n^1 have n in common.

- mathstudent55

That means the GCF of the two terms is 5m^2n

- baby456

how did you get 2 from

- mathstudent55

The common factor is
\(\large 5m^2n\)
The 2 is the exponent of the m.

- baby456

ok

- mathstudent55

\(\large 10m^3n^2-15m^2n^1\)
Now we factor out the GCF:
\(\large =5m^2n(~~~~~~~~~~~~~~)\)

- baby456

how? i dont know i am so confused

- baby456

what gcf

- mathstudent55

\(\large =5m^2n(2mn - 3)\)

- baby456

where are you getting the 2 and 3

- mathstudent55

The 2 and the 3 are the numbers you need so that when you multiply by the 5 outside you get back the 10 and 15 you started with.

- mathstudent55

Factoring is the opposite of the distributive property.

- baby456

i dont get it

- mathstudent55

Let me go back a few steps and show a simple example.
Do you know the distributive property?

- baby456

what?

- mathstudent55

Use the distributive property to simplify this expression:
\(2(3 + 5)\)

- baby456

11 i think

- mathstudent55

No.
I'll show you.
The distributive property of multiplication over addition is this:
\(2(3 + 5) = 2 \times 3 + 2 \times 5\)

- mathstudent55

|dw:1435095136526:dw|

- mathstudent55

The distributive property shows you that you multiply the number outside parentheses by each of the numbers inside the parentheses. The you add the products.

- mathstudent55

Here is another example:
|dw:1435095244847:dw|

- baby456

ok

- baby456

but how does that help with factoring.

- mathstudent55

Factoring is doing the distributive property in reverse.

- mathstudent55

Here is an example of the distributive property using variables.
|dw:1435095365764:dw|

- baby456

ok

- mathstudent55

We went from a product of a variable, m, and a binomial, 2m + n,
to a result, \(2m^2 + mn\)

- mathstudent55

Now let's do that problem in reverse.
Let's say we are given the final answer above, and we are asked to factor.

- mathstudent55

This is the problem we have now:
|dw:1435095505073:dw|

- mathstudent55

First, we find the greatest common factor of the two terms.

- mathstudent55

|dw:1435095568553:dw|

- mathstudent55

Now we take the problem, and we do the distributive property in reverse.

- mathstudent55

We need to find what to place inside the parentheses to get back to our original problem.
|dw:1435095632204:dw|

- mathstudent55

|dw:1435095685363:dw|

- mathstudent55

|dw:1435095737610:dw|

- mathstudent55

|dw:1435095761968:dw|

- mathstudent55

|dw:1435095800516:dw|

- baby456

i dont get it. it like my brain is stuck.

- mathstudent55

This is the answer:
|dw:1435095820650:dw|

- mathstudent55

The way you check the answer is to multiply it out using the distributive property to make sure you get back to the given problem.

- baby456

where did you get THAT HOW PROB ELM THE 2M THING.

- mathstudent55

Go over this a few times. Sometimes after you go over something a few times you begin to understand.

- mathstudent55

In the problem I made upi above, which is to factor
\(2m^2 + mn\)
Did you understand how I got the GCF of m?

- mathstudent55

|dw:1435096046142:dw|