Factoring PolynomialsJune 19, 2009
For some reason, I always seem to freeze up when I need to factor a polynomial where a>1. I don’t know why that didn’t stick with me the way multiplying binomials did or factoring when a=1. So, here’re the two methods I know:
1st Method (grouping)
- Find two numbers whose sum is the middle term and whose product is the product of the first and last terms.
- Rewrite the expression so that you split the middle term into two terms using the numbers found in step 1.
- Group the first two terms together and factor them; then group the second two terms and factor those.
- Use the distributive property to write the expression as two binomials.
2nd Method (My Father Drives a Red Mustang)
- First, factor out any common terms.
- Multiply – Multiply the first and last coefficients together.
- Factor – As before, find two numbers whose sum is the second coefficient and whose product is the number found in the previous step. Set up two sets of parentheses.
- Divide – Divide both numbers by the first coefficient.
- Reduce – Reduce any fractions if possible.
- Move – Move the remaining denominators in front of the variable(s).
Example: Factor .
- Find two numbers whose sum is -7 and whose product is -30. This would be -10 and 3.
- The expression then becomes (because ).
- Grouping and factoring produces .
- Reverse distributive property: .
- The expression is already reduced.
- My –
- Father – As before, we get -10 and 3.
- Drives a – Set up the parentheses and divide:
- Red – Reducing the fractions gives us:
- Mustang – Move the denominator:
Now I have to say that the second method, while it works, bothered me until I figured out the reason why it works. This isn’t a formal proof, but I understand it anyway:
Consider the expression . It would multiply to the expression . Factoring this a bit gives us . Using the “Mustang” method to factor this, we can easily see that the two numbers that add up to and multiply to would be and . Therefore, our expression would look like: . Reducing this gives us: . Moving the denominators gives us our original expression.
Now, let’s consider the original binomial product expression: . If we factored out of the first binomial and out of the second binomial, we would get: or . This is why this method works — we’re not really moving the denominator, so much as factoring it out and then remultiplying by it again. Rewriting the example’s last three steps would then be: