Book 1 Algebra Chapter 3 Q 20 - Pg70

[leah.coiro]

Hello,

Could someone help me with the steps to solve this problem? (It's Q20 in the check your skills section, chapter 3, book 1, 4th edition.)

Problem: 4a²+4ab+b²=0
Solution: (2a+b²)=0

The steps taken in the answer section aren't clicking with me. I don't understand HOW this is factored the way it is in the book. Guidance?

Thanks!

[tommywallach]

No problem! (Though you put your squared symbol in the wrong place in the solution you posted!)

However, I would encourage you NOT to think of this in steps, but as something you should (in future) be able to immediately recognize.

When you see this: 4a²+4ab+b²

You should immediately think of the quadratic identity you know: a²+2ab+b²

You should know that this represents: (a+b)²

So the question then becomes: How do we get a 4 in front of that a^2 and a 4 in front of the ab, but NO 4 in front of the b^2.

At that point you KNOW that whatever ends up in the parentheses, it can ONLY be a lone b (because that keeps the b^2). So the question becomes: What do I put in front of the a?

At that point, it should be obvious, because you're squaring. You have to put a 2 in front of the a, so that you'll end up with a 4a^2. The middle term (4ab) just sorts itself out in the wash, more or less.

Does that make more sense? I'm not trying to say this is easy, only that it's easier to think of it in terms of logic, instead of some kind of multi-step simplification.

-t