HOW DOES???

[dapplegate]

Study Guide 1: PG 94
X^2-5X-X=0
become
(X-6)(X+1)=0


AND

Study Guide 1: PG 102
n^2-14n+49=0
become
(n-7)(n-7)=0

[jnelson0612]

Study Guide 1: PG 94
X^2-5X-6=0
become
(X-6)(X+1)=0


AND

Study Guide 1: PG 102
n^2-14n+49=0
become
(n-7)(n-7)=0

dapplegate,
Let's examine the first example you provided (I corrected a typo):
X^2 - 5X - 6 = 0.

In the case of quadratics, remember that we obtained this expression by multiplying together two expressions similar to this: (X +/- number)(X +/- number)=0.

We use FOIL to multiply these expressions together to obtain the original quadratic. FOIL standards for First, Outside, Inside, Last. We multiply the first terms together (X * X), then the two outside terms (X * number), then the two inside terms (number * X), then the last terms (number * number).

Since our quadratic contains X^2, we must FIRST multiply together X and X. Thus we know we have (X +/- number)(X +/- number)=0

Here is the tricky part. We need to multiply two numbers together that give us -6. We must be multiplying a positive and a negative because we obtain a negative. Our two pairs of factors for 6 are 1,6 and 2,3.

Whatever pair of numbers we select here will also be multiplied by X and X when I multiply the "outside" and "inside" terms. This result must result in -5X.

When I test the pairs I realize that I must use 1 and -6. That results in -6 when multiplied together, and when each is multiplied by X I obtain X + (-6X) which equals -5X. We have now obtained the original quadratic equation.

I realize this is a long and possibly confusing explanation. You can learn a lot more about factoring quadratics in our Foundations of Math book.

Thank you,