So... who here knows how to factor the differences of squares?

[Esque]

Algebra 1. The teacher isn't at school, and the sub is like, "I read book. You do math. No yelling. kthxbai." Google was unhelpful. So... do any of you know how to factor the differences of squares? I do psychology and English, not Algebra... :sweatdrop: Do you need a problem?

Directions: Apply a factoring technique more than once.
Problem: 6x^3+30x^2-24x-120

That's the one with the smallest amount of numbers.

 

[1. Luftballon]

Algebra 1. The teacher isn't at school, and the sub is like, "I read book. You do math. No yelling. kthxbai." Google was unhelpful. So... do any of you know how to factor the differences of squares? I do psychology and English, not Algebra... :sweatdrop: Do you need a problem?

Directions: Apply a factoring technique more than once.
Problem: 6x^3+30x^2-24x-120

That's the one with the smallest amount of numbers.

6x^3+30x^2-24x-120 = 6(x^3+5x^2-4x-20)

uh...

 

[Diz]

http://www.wolframalpha.com/input/?i=factor+6x^3+30x^2-24x-120

Wolfram Alpha is a really good site for doing math stuff.

The answer in this case would be 6 (x-2) (x+2) (x+5)

 

[opaltiger]

Generally:

(a^2 - b^2) = (a - b)(a + b)

Which does seem to appear somewhere in the result, but first you have to reduce the polynomial to a quadratic expression.

 

[Diz]

Our school has simplified it into an acronym, FOIL. FOIL stands for First, Outside, Inside and Last

Say you are solving a problem like what Opaltiger mentioned. (a - b)(a + b)
You multiply the first terms, 'a' and 'a' to get 'a^2'. Then you multiply the out side terms, 'a' and 'b' to get 'ab'. Then multiply in side terms a 'negative b' and an 'a', resulting in '-ab'. Finally, the last terms, 'negative b' and 'positive b. We know that a negative number times a positive number gets us a negative number so it is a '-b^2'.

Now you put this all together 'a^2 + ab - ab - b^2'. Because the middle two terms are similar, we can combine them. And in this case they cancel each other out because they are the same absolute value, but with different signs. So we end up with 'a^2-b^2'.

Note: In most problems 'a' will be a variable, with or without a coefficient, and 'b' will be a normal number.

 

[Kali the Flygon]

When you're dealing with those polynomials of order 3, and know that they're supposed to have a difference of squares, you should look for a common ration between the x^3 coefficient and x^2 coefficient, and between the x and 1 coefficient. That is, if you're factoring a*x^3 + b*x^2 - c*x - d, check the ratios b/a and d/c. If they are the same, you know you can use the difference of squares. You can factor out a(x + b/a) from the polynomial, leaving yourself with x^2 - (c/a), which factors into (x + sqrt(c/a)) and (x - sqrt(c/a)).

 

[Esque]

I think maybe I should explain this better. I think I understand what I'm supposed to be doing, and I know (vaguely) how to factor. However, in all of the example problems of gotten on this subject, the two sides of the equals sign look like completely different and unrelated problems. a^2-b^2 doesn't have anything to do with a + b!

 

[Kali the Flygon]

a + b might not look like a^2 - b^2, but it will always be a factor of it. I don't quite understand where you're getting stuck.

 

[Togetic]

Okay...Lets do this the long way.

a^2 - b^2

(a - b)(a + b)

a x a = a^2
a x b = ab
-b x a = -ab
-b x b = -b^2

Since ab & -ab cancel out, it means that (a - b)(a + b)
is equal to a^2 - b ^2

 

[Esque]

Oh. That makes sense. Thank you.