## anonymous one year ago Can someone help me with an algerba 1 question?

1. diamondboy

Sure

2. anonymous

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

3. diamondboy

What's d question?

4. anonymous

so bassicaly find the GCF

5. diamondboy

ok

6. mathstudent55

$$2x^3y^4 − 8x^2y^3 + 6xy^2$$

7. anonymous

yes

8. 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

9. mathstudent55

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

10. diamondboy

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11. diamondboy

can you go on from this?

12. mathstudent55

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

13. anonymous

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

14. diamondboy

yes

15. mathstudent55

|dw:1442526095341:dw|

16. mathstudent55

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

17. anonymous

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

18. anonymous

3 and 2

19. 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. $$\color{red}2\cdot x^3 \cdot y^4 − \color{red}2\cdot 2 \cdot 2 \cdot x^2y^3 + \color{red}2 \cdot 3 \cdot xy^2$$

20. anonymous

yes i understand that :)

21. 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.

22. anonymous

okay I got that

23. anonymous

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

24. mathstudent55

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

25. 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.

26. anonymous

so 2x? is what your dividing everything by?

27. anonymous

or 2xy?

28. anonymous

2xy^2

29. 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.

30. mathstudent55

Correct. Use 2xy^2.

31. anonymous

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

32. mathstudent55

ok

33. anonymous

ok heres what i got

34. anonymous

2xy2(x2y2 − 4xy + 3)

35. mathstudent55

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

36. mathstudent55

We are not done, though.

37. anonymous

AHhhh lol i thought we were

38. mathstudent55

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

39. anonymous

yes

40. 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.

41. anonymous

ok

42. mathstudent55

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

43. mathstudent55

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

44. anonymous

ok

45. 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.

46. anonymous

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

47. mathstudent55

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

48. anonymous

Thank you so much! :D

49. mathstudent55

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

50. anonymous

Ok

51. 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: $$=2xy^2(xy -3)(xy - 1)$$

52. mathstudent55

You're welcome.