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Date: 7/22/96 at 23:38:0
From: Anonymous

Hello,

Can you explain in a simple form, how to do factoring? Following are
some examples of expressions I need to factor:

a).  3y^2 - 3

b).  y^2 + 4y -12

c).  6x^2 + 30x - 900

d).  4x^2 + 24x + 36

Are there common points or rules I can apply in every situation?

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Date: 7/23/96 at 10:4:55
From: Doctor Paul

a) I think the idea in factoring is to look for a number or a variable
that is present in every term of what needs factoring.
For example, in:

3y^2 - 3

the only thing that can be factored here is the 3 that is present in
each term.  Factor the 3 out to the front and you get:

3 * ( y^2 - 1)

b) When factoring quadratic equations there generally is not anything
that is common in every term. In this case we cannot simply pull a
constant out front and rewrite the problem. What we *can* do is
this... you should know that your answer will look like this form:

(y + a) * (y + b)

The trick is knowing how to find a and b. Here's how to do it for the
following equation:

y^2 + 4y + -12

Take the term with no y's attached to it (the constant term). In this
particular case, the constant term is the -12.  Divide the -12 into
what I call groups. Each group is a set of two numbers that when
multiplied together will give -12. With the number -12, I can think of
these groups:

{-1,12}, {-2,6}, {-3,4}, {1,-12}, {2,-6}, {3,-4}

Now you have six different groups. The next thing to do is to find out
what the coefficient of the 'y' term is. That is, we want to know
what number is being multiplied by 'y' in the quadratic equation. In
this case that number is 4, right?  What we want to do is add the two
numbers of each group together and find out which one will give us 4.
If you add -1 and 12 you get 11. That's not right. So we try again:
if you add -2 and 6 then you get 4! That is the group we want.

Feel free to add the rest of the groups. You will not get 4 again.

Here is what is really important:

The numbers in the group you chose (in this case -2 and 6) will be
the a and the b that we have been looking for. Which one you chose for
a and which you chose for b does not matter. I will just use  a = -2
and b = 6,

(y + (-2) * (y + 6)

= (y - 2)(y + 6)

If you multiply it out and you get what you started with then that is

Parts c and d require you to first divide by a constant and then
equations you always want to divide by the coefficient (or number)
that is multiplied by the y^2 term. So for part c you would divide
everything by 6, right? Then factor as I showed you for part b.

Good luck.

-Doctor Paul,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra

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