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 Skip Navigation LinksMath Help > Basic Algebra > Factoring > Very Basic Factoring

Very Basic Factoring

On 9/15/01 11:29:45 AM, Anonymous wrote:
>Can you show all the steps to
>these 2 problems and explain
>what's going on.
>
>k^2+8k+7

In answering your question, I write the way I would explain it to you. So read this slowly, and re-read it a few times to understand it.

The trick to reversing the FOIL operation for this problem is to realize that the answer will look like this:

(k+___) (k+___)

When you FOIL the expression, above, the sum of the two numbers in the blanks will be 8, because the "O" and "I" are each a number in one of the blanks multiplied by k. Their sum is 8k.

Also, when you FOIL the expression, above, the product of the two numbers in the blanks will b 7, because the "L" is the product of those two numbers.

So you're looking for a pair of numbers whose product is 7 and whose sum is 8. It's not hard to find them, because there aren't very many factors of 7, being that 7 is a prime number. In fact the only factors of 7 are 1 and 7, and lo and behold, their sum is 8.

So the numbers in the two blanks are 1 and 7. The answer is

(k+1)(k+7)

>m^2-10mn+21n^2

In this problem you have two variables. That makes it look tricky, but I want you to notice something: in each term, whenever the power of m is higher, the power of n is lower, and vice versa. That means the answer will look like this:

(m - ___n) (m - ___n)

Oh, and why the two minus signs instead of plus signs? Because the middle term -- the sum -- is negative, and the last term -- the product -- is positive. Two numbers must both be negative if their sum is negative and their product is positive.

Look once more at the two factors, above, with the blanks in them. Each of the operations in FOIL takes two variables: either two m's, two n's or one of each. Take each one separately: F takes two m's, O and I each take one m and one n, and L takes two n's. As you can see whenever the power of m increases, the power of n decreases, and vice versa.

I've been "talking" long enough. It's time to get back to the problem.

m^2-10mn+21n^2

(m - ___n) (m - ___n)

Now, as with the previous problem, we're looking for numbers to put in the blanks, two negative numbers. Their sum is -10, and their product is 21. If the two numbers jumped into your head, great. If not, don't worry. Make a list of factors of 21, and organize them in pairs so that the product of each pair of factors is 21. It looks like this:

-1 * -21 = 21
-3 * -7 = 21

So there are two possibilities for the numbers in the blanks. But only one of those possibilities is a pair of numbers whose sum is -10.

(m - 3n) (m - 7n)

Reversing FOIL isn't easy or natural, so don't worry if you have trouble with it. It takes LOTS and LOTS of practice to do these problems.

Related pages in this website

Next: If you understand this page, and you'd like to go on to more advanced factoring, click here.

 

The webmaster and author of this Math Help site is Graeme McRae.