SOLUTION: I need help factoring some equations please help: 2. 16xy + 32yz � 20xyz 3. 9x2 � 64 4. 5x3 � 125u2x 5. x2 + 7x + 12 Please explain to me how to do these because I am co

Question 211437: I need help factoring some equations please help:
2. 16xy + 32yz � 20xyz
3. 9x2 � 64
4. 5x3 � 125u2x
5. x2 + 7x + 12
Please explain to me how to do these because I am confused and old.
Found 2 solutions by Alan3354, drj:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
2. 16xy + 32yz � 20xyz
All terms are even numbers and have a y
4y*(4x + 8z - 5xz)
That's all that can be done.
---------------
3. 9x2 � 64
Any difference of 2 squares is a plus and minus situation
(3x + 8)*(3x - 8)
-------------------
4. 5x3 � 125u2x
Factor out the 5x
= 5x*(x^2 - 25u^2)
Then there's difference to 2 squares
= 5x*(x - 5u)*(x + 5u)
--------------
5. x2 + 7x + 12
For this one, look for factors of 12 that add to 7.
12 = 1*12 --> 13 NG
12 = 2*6 --> 8 NG
12 = 3*4 -- 7 That's it
= (x+3)*(x+4)

Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
I need help factoring some equations please help:
2. 16xy + 32yz � 20xyz
3. 9x2 � 64
4. 5x3 � 125u2x
5. x2 + 7x + 12
Please explain to me how to do these because I am confused and old.

2. 16xy+32yz-20xyz. Take a look at all three terms and find what is common. That is, find the common factor.

2a. In this case, let's start with the common factor for the numbers 16, 32, -20. The common factor is 4. How did I calculate this? Find the lowest common factors: 16=2*2*2*2, 32=2*2*2*2*2, -20=-2*2*5. The common factor is 2*2=4. So factor out 4 in all three terms which yields: 4(4xy+8yz-5xyz)

2b. Now, check if x is common in all three terms. x is not a common factor. So it cannot be factored out.

2c. Now check if y is common to all three terms. y is a common factor. So factor out y. 4y(4x+8z-5xz).

2d. Now check if z is common to all three terms. z is not a common factor.

2e. Therefore, final answer is 4y(4x+8z-5xz).

3. 9x2 � 64.

3a. This is a special case of the FOIL method. That is, multiply (Ax-B)(Ax+B) where A=3 or Ax=3x and B=8. FOIL means F=First O=Outer, I=Inner, L=Last: Multiply First Terms (Ax*Ax), then Outer Terms, Ax*B, then Inner Terms, -B*Ax, then Last Terms, -B*B. Therefore,

3b. (Ax-B)(Ax+B)=Ax*Ax+Ax*B-B*Ax-B*B. We can simplify this since the INNER and OUTER terms add up to zero. Simplifying, yields (Ax-B)(Ax+B)=(Ax)^2-B^2.

3c. So 9x2-64=(3x-8)(3x+8) is the final answer where A=3 or Ax=3x and B=8.

4. 5x3 � 125u2x. Follows a similar process as Problem 2.

4a. 5x3-125u2x= 5(x3-25u2x). Factored common factor 5.
4b. 5(x3-25u2x)=5x(x2-25u2). Factors common factor x. 5x(x2-25u2) can be rewritten as 5x(x^2-25u^2), where u2 is meant as u-squared or u^2.

5. x2 + 7x + 12 x2+7x+12.

5a. Based on the FOIL method again, see above Problem 3. You need to satisfy two equations: A+B=7 and A*B=12. So you need to find factors of 12 that add to seven. You can use a trial and error method for this case.

5b. 2 and 6 are factors of twelve but they don't add to seven.
1 and 12 are factors of twelve but they don't add to seven.

5c. Now, let's try 3 and 4. When you multiply these numbers, you get 3*4=12. When you add these numbers, you get 3+4=7.

5d. Therefore, x2+7x+12=(x+3)(x+4) is the answer. Now, use FOIL method to check answer:

(x+3)(x+4)=x*x+4x+3x+3*4=x2+7x+12. So it works.

Note: If you prefer a visual approach to factoring, I have developed free step-by-step videos (1-2 hours) on Factoring at http://www.FreedomUniversity.TV/courses/IntroAlgebra/Module5.html. The set of videos have many similar problems as the ones that are describe above.

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