Can someone help me with an algerba 1 question?

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Can someone help me with an algerba 1 question?

Mathematics
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Sure
The question is Factor completely 2x3y4 − 8x2y3 + 6xy2.
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so bassicaly find the GCF
ok
\(2x^3y^4 − 8x^2y^3 + 6xy^2\)
yes
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
Step 1 of factoring: Try to factor a common factor. What is the largest common factor of all three terms?
|dw:1442526155297:dw|
can you go on from this?
Correct. The number part of the GCF is 2. Now you need the variable part of the GCF.
Okay it said start with a varible with the lowest degree i dont know what that means is the the exponent?
yes
|dw:1442526095341:dw|
Degree of a variable is the exponent. You are correct.
okay but both x and y have an exponet that is the same
3 and 2
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. \(\color{red}2\cdot x^3 \cdot y^4 − \color{red}2\cdot 2 \cdot 2 \cdot x^2y^3 + \color{red}2 \cdot 3 \cdot xy^2\)
yes i understand that :)
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.
okay I got that
so which ever one has the less is the one you need?
\(\color{red}2\cdot \color{green}x \cdot x \cdot x \cdot \color{purple}{y \cdot y} \cdot y \cdot y − \color{red}2\cdot 2 \cdot 2 \cdot \color{green}x \cdot x\cdot \color{purple}{y \cdot y} \cdot y + \color{red}2 \cdot 3 \cdot \color{green}x \cdot \color{purple}{y \cdot y}\)
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.
so 2x? is what your dividing everything by?
or 2xy?
2xy^2
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.
Correct. Use 2xy^2.
okay imma try to see if i can fiqure i out and tell me if im right?
ok
ok heres what i got
2xy2(x2y2 − 4xy + 3)
\(2xy^2(x^2y^2 - 4 xy + 3)\) Excellent. You got the same I did.
We are not done, though.
AHhhh lol i thought we were
Now think of the product xy as a variable. Let's say xy = z. That means x^2y^2 = z^2
yes
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.
ok
Now we have this: \(=2xy^2(x^2y^2 - 4 xy + 3)\) After the substitution we have this: \(=2xy^2(z^2 - 4 z + 3)\)
The trinomial in the parentheses is a quadratic trinomial in z. We need to try to factor it.
ok
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.
Also i forgot to post the answers. 2(x3y4 − 4x2y3 + 3xy2) 2x(x2y4 − 4xy3 + 3y2) 2xy2(x2y2 − 4xy + 3) Prime
According to your choices, we are done, and the answer is C.
Thank you so much! :D
Just for completeness, let's just finish factoring because the trinomial is also factorable.
Ok
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: \(=2xy^2(xy -3)(xy - 1)\)
You're welcome.

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