Can someone help me with an algerba 1 question?

- anonymous

Can someone help me with an algerba 1 question?

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- diamondboy

Sure

- anonymous

The question is Factor completely 2x3y4 − 8x2y3 + 6xy2.

- diamondboy

What's d question?

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## More answers

- anonymous

so bassicaly find the GCF

- diamondboy

ok

- mathstudent55

\(2x^3y^4 − 8x^2y^3 + 6xy^2\)

- anonymous

yes

- anonymous

The notes that i took in class says that first i need to find a number that goes into all three coeffiecent evenly which is 2

- mathstudent55

Step 1 of factoring:
Try to factor a common factor.
What is the largest common factor of all three terms?

- diamondboy

|dw:1442526155297:dw|

- diamondboy

can you go on from this?

- mathstudent55

Correct. The number part of the GCF is 2.
Now you need the variable part of the GCF.

- anonymous

Okay it said start with a varible with the lowest degree i dont know what that means
is the the exponent?

- diamondboy

yes

- mathstudent55

|dw:1442526095341:dw|

- mathstudent55

Degree of a variable is the exponent. You are correct.

- anonymous

okay but both x and y have an exponet that is the same

- anonymous

3 and 2

- mathstudent55

Below, I broke up the numbers into their factors.
You can see clearly (the red helps) that the only number factor all three terms have is 2.
\(\color{red}2\cdot x^3 \cdot y^4 − \color{red}2\cdot 2 \cdot 2 \cdot x^2y^3 + \color{red}2 \cdot 3 \cdot xy^2\)

- anonymous

yes i understand that :)

- mathstudent55

Now let's look at the variables to see what variables appear in all factors.
\(\color{red}2\cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y − \color{red}2\cdot 2 \cdot 2 \cdot x \cdot x\cdot y \cdot y \cdot y + \color{red}2 \cdot 3 \cdot x \cdot y \cdot y\)
Normally, you don;t have to do what I did above, but I just want you to see what the exponents really mean.
Now we see how many x's and y's are in common to all terms.

- anonymous

okay I got that

- anonymous

so which ever one has the less is the one you need?

- mathstudent55

\(\color{red}2\cdot \color{green}x \cdot x \cdot x \cdot \color{purple}{y \cdot y} \cdot y \cdot y − \color{red}2\cdot 2 \cdot 2 \cdot \color{green}x \cdot x\cdot \color{purple}{y \cdot y} \cdot y + \color{red}2 \cdot 3 \cdot \color{green}x \cdot \color{purple}{y \cdot y}\)

- mathstudent55

Yes.
Since the 3rd term only has one x, and the other terms have more than one x, the three terms have only 1 x in common.
Since the third term has only 2 y's, and the other terms have more than 2 y's, all three terms have only 2 y's in common.

- anonymous

so 2x? is what your dividing everything by?

- anonymous

or 2xy?

- anonymous

2xy^2

- mathstudent55

Every term has a factor of 2.
Every term has a factor of x
Every term has a factor of y^2
The GCF is 2xy^2
We divide all three terms by 2xy^2
Now we pull out the common factor, 2xy^2.

- mathstudent55

Correct. Use 2xy^2.

- anonymous

okay imma try to see if i can fiqure i out and tell me if im right?

- mathstudent55

ok

- anonymous

ok heres what i got

- anonymous

2xy2(x2y2 − 4xy + 3)

- mathstudent55

\(2xy^2(x^2y^2 - 4 xy + 3)\)
Excellent. You got the same I did.

- mathstudent55

We are not done, though.

- anonymous

AHhhh lol
i thought we were

- mathstudent55

Now think of the product xy as a variable.
Let's say xy = z.
That means x^2y^2 = z^2

- anonymous

yes

- mathstudent55

If you think of the variables that way, we can make a substitution. Wherever we see xy, we will replace it with z. Then wherever we see x^2y^2 we will replace it with z^2.

- anonymous

ok

- mathstudent55

Now we have this:
\(=2xy^2(x^2y^2 - 4 xy + 3)\)
After the substitution we have this:
\(=2xy^2(z^2 - 4 z + 3)\)

- mathstudent55

The trinomial in the parentheses is a quadratic trinomial in z.
We need to try to factor it.

- anonymous

ok

- mathstudent55

To factor a trinomial of the form
\(x^2 + ax + c\),
you need to find two number that multiply to b and add to a.
Our trinomial is of that form, so wee need to try to find two numbers that multiply to 3 and add to -4.

- anonymous

Also i forgot to post the answers.
2(x3y4 − 4x2y3 + 3xy2)
2x(x2y4 − 4xy3 + 3y2)
2xy2(x2y2 − 4xy + 3)
Prime

- mathstudent55

According to your choices, we are done, and the answer is C.

- anonymous

Thank you so much! :D

- mathstudent55

Just for completeness, let's just finish factoring because the trinomial is also factorable.

- anonymous

Ok

- mathstudent55

From now on, it's just extra info. In fact, from the point I began with the substitution it's extra info.
Since the two numbers have to multiply to 3, a positive number, the two numbers must be both positive or both negative.
Since they need to add to -4, a negative number, the two numbers must contain at least one negative number.
The two numbers are -3 and -1
\(=2xy^2(z -3)(z - 1)\)
Now we substitute xy back for z to get the final complete factorization:
\(=2xy^2(xy -3)(xy - 1)\)

- mathstudent55

You're welcome.

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