Why is it a good idea to factor out the GCF first and then use other methods of factoring?

Factoring out the 3 and giving you 3(x^2-6x+5) decreases the amount of time and effort you have to use in trial and error to find the factors which can easily be seen now to be (x+6)(x-1)

How would I solve this equation using factoring and show my work?

I made this equation to help you: w=width *w(2+w)=24 that would mean... *w^2 + 2w =24 from there I just guessed and found the answer to be 4

How much would you guess a factoring company would purchase 2 billion dollars of debt for?

You'd have to know the terms of the loans -- e,g, interest rate, term, payment terms, etc,

Can I use a factoring company to buy a profitable business ?

You are talking about buying a business based upon its receiveables. Borrowing against the receivables to finance the business. I highly doubt it. Keep in mind factoring is going to be like getting 70% of the value of the invoice.It might be possible, call a factoring company, be prepared to have a resume that shows why you can run this company. Good Luck

What is the formula you follow when factoring the difference of two squares?

Factor out the greatest common factor If the parentheses don't match then it can't be done that way. Maybe if you gave me an example, I could show you.

What are some techniques for factoring polynomials of degree higher than 2?

There are three main ways to factor equations with degree higher than 2: 1. For cubics there are rules for factoring the sum or difference of cubes. Memorize them. 2. For relatively small degrees above 2, grouping can sometimes be used. 3. The most general method is to look at the graph or use the rational roots theorem to find at least one root, convert the root(s) just found into factor(s) using the factor theorem, then divide the original function by those factor(s). Then perhaps apply the method again to the reduced function.

FACTORING ....!?!??!?!?

72(m-9/8)(m+8/9) The factorization of ax^2 + bx + c is a(x-x1)(x-x2), where x1, x2 are the roots of ax^2 + bx + c = 0

FACTORiNG??????

when you have a difference of squares (x^2 - y^2) this is equal to (x+y)(x-y) therefore 1. 100x^2 - 64y^2 = (10x+8y)(10x-8y) another technique is expanding and grouping in the equation y=ax^2 + by + c, you multiply a and c together. Then you find factors of ac that add up to b 2. n^2 - 5n - 36 1*-36=-36 factors that add up to -5 are -9 and 4 therefore, you can also say n^2 - 9n + 4n - 36 this does not effect the value, so we are okay. Now you group this by working with the first two and last two terms separately. First two: n^2 - 9n = n(n-9) Last two: 4n - 36 = 4(n-9) therefore, the whole thing is n(n-9) + 4(n-9) the fact that (n-9) shows up there twice , which means we are okay now we can distribute out the (n-9) (n-9)(n+4) when it is possible to distribute out a factor initially, go ahead and do so. 3. 2a^3 + 16a^2 + 30a 2a(a^2 + 8a + 15) all of a sudden, this is so much easier to factor. Facotrs of 15 that add up to 8 are 5 and 3 you end up with 2a(a+3)(a+5) 4. 6x^2 + 20x - 16 2 (3x^2 + 10x - 8) 2 (3x^2 -2x + 12x - 8) 3 [x(3x-2) + 4(3x-2)] 3(x+4)(3x-2) 5. 9x^2 + 12x + 4 9x^2 + 6x + 6x + 4 3x(3x+2) + 2(3x+2) (3x+2) ^2 I did this work, but by looking at the first and last terms, you know they are both perfect squares, so there was a pretty good chance that the whole polynomial was a squared binomial....it's a bit complicated, so just use the expansion and grouping method.

factoring.....?

Here is a great tool to help you step by step through any factoring problem you have (you can plug in any problem and it will show you how it's done. It's like a calculator that can factor) http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/quadratic-factoring.solver