# Distributive Multiplication to factor common Polynomials?

In a math class I am taking, we are discussing how to factor polynomials with grouping. For the most part, I've understood everything so far, until I was told that I needed to "Use the Distributive Law of Multiplication to factor the common polynomial out of each term".

Original: $$x^3 - 6x^2 - 3x + 18$$

In the example, after a GCF is determined as -3. At this point, factoring those out results in: $$(x^3 - 6x^2) + (-3x + 18) = x^2(x - 6) - 3(x - 6)$$

Then comes the part where it says to use the Distributive Law of Multiplication which is confusing me. This is how the example works out: $$x^2(x-6) - 3(x-6) = (x-6)(x^2-3)$$

From what I understand, distributive would mean I'd take the x squared value and multiply it by both values inside the parenthesis (x and -6) but that doesn't result in (x-6) as the example suggests. Or does it? I'm really struggling to understand what is occuring here. I've always been under the impression that the distributive property works like: $$a(b+c) = ab + ac$$

In my mind, I would simplify the above as: $$x^2(x-6) - 3(x-6)$$ $$x^3-6x^2 - 3x - 18$$

But that's not the answer :-/

Would anyone with a moment to spare care to enlighten me on how this is working?

Look at it this way...

For the general case you have a(b+c)=ab+ac.

Now let a=(x-6), b=x^2 and c=3

So for this particular case you have (x−6)(x^2−3)

You distribute the "a" term across the sum of b and c so you get...

(x−6)x^2−(x−6)3

For the work you are doing, you are actually trying to do this in reverse. If you have a common term such as (x-6) distributed across all other terms in your polynomial, you can factor this term out which basically undistributes the common term. The goal is to break the polynomial into smaller parts so you are more likely to be able to find the factors.

• OOOOhhhhhh. I stared at your answer for a few minutes trying to understand what you meant and then a lightbulb clicked on! Now this makes sense. Thank you. – Jeremy Harris Jan 14 '13 at 15:01

You're right about distributivity, you're just applying it in the wrong direction – if you start with the right-hand side, everything will make sense. The order of the factors has been switched; it may be easier to see distributivity at work if you swap them.

• Thank you for your response, I'm not sure I follow it though. Can you elaborate a little more? There is something I'm fundamentally not grasping and I haven't had that "aha" moment yet. – Jeremy Harris Jan 14 '13 at 14:57
• @cillosis: I see your comment under Bruce's answer is more recent than this one -- everything cleared up? – joriki Jan 14 '13 at 16:32

You are correct about the distributive property: it says that $a(b+c)=ab+ac$. In your case, you use it with $a=x-6$, $b=x^2$ and $c=-3$. You get $$(x-6)(x^2-3) = (x-6)x^2 + (x-6)(-3) = x^2(x-6) - 3(x-6).$$