## Factoring Polynomials

June 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)

1. Find two numbers whose sum is the middle term and whose product is the product of the first and last terms.
2. Rewrite the expression so that you split the middle term into two terms using the numbers found in step 1.
3. Group the first two terms together and factor them; then group the second two terms and factor those.
4. 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 $5x^2-7x-6$.
1st Method

1. Find two numbers whose sum is -7 and whose product is -30. This would be -10 and 3.
2. The expression then becomes $5x^2-10x+3x-6$ (because $-10x+3x=-7x$).
3. Grouping and factoring produces $5x(x-2)+3(x-2)$.
4. Reverse distributive property: $(5x+3)(x-2)$.

2nd Method

• The expression is already reduced.
• My$(5)(-6)=-30$
• Father – As before, we get -10 and 3.
• Drives a – Set up the parentheses and divide: $(x+\frac{-10}{5})(x+\frac{3}{5})$
• Red – Reducing the fractions gives us: $(x-2)(x+\frac{3}{5})$
• Mustang – Move the denominator: $(x-2)(5x+3)$
• 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 $(ax+b)(cx+d)$. It would multiply to the expression $acx^2+bcx+adx+bd$. Factoring this a bit gives us $acx^2+(bc+ad)x+bd$. Using the “Mustang” method to factor this, we can easily see that the two numbers that add up to $(bc+ad)$ and multiply to $(ac)(bd)$ would be $bc$ and $ad$. Therefore, our expression would look like: $(x+\frac{bc}{ac})(x+\frac{ad}{ac})$. Reducing this gives us: $(x+\frac{b}{a})(x+\frac{d}{c})$. Moving the denominators gives us our original expression.

Now, let’s consider the original binomial product expression: $(ax+b)(cx+d)$. If we factored $a$ out of the first binomial and $c$ out of the second binomial, we would get: $[a(x+\frac{b}{a})][c(x+\frac{d}{c})]$ or $ac[(x+\frac{b}{a})(x+\frac{d}{c})]$. 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:

$5[(x+\frac{-10}{5})(x+\frac{3}{5})]$
$5[(x-2)(x+\frac{3}{5})]$
$(x-2)(5x+3)$