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Video: Factorizing Perfect Square Trinomials

Tim Burnham

Factor 𝑎² βˆ’ 6𝑎𝑏 + 9𝑏².

04:38

Video Transcript

Factor π‘Ž squared minus six π‘Žπ‘ plus nine 𝑏 squared.

Now in this expression, we’ve got one π‘Ž squared β€” I’m gonna put that in there β€” and we’ve got nine 𝑏 squared, and then we’ve got this term in the middle which is six times π‘Ž times 𝑏.

Now the fact that the highest power is two and we’ve got this particular pattern means that we’re gonna two pairs of parentheses like this that we can multiply together to make up that expression. So in the first parenthesis, we’re gonna have something plus or minus something; and in the second one, we’re gonna have something plus or minus something else.

Now let’s think about this in reverse if I had some parentheses like this and I was going to multiply them together, I would have to multiply this term by this term and this term and this term by this term and this term. Now the fact that I’ve got one π‘Ž squared means I could put in π‘Ž here and an π‘Ž here and when I multiply π‘Ž by π‘Ž I’m gonna π‘Ž squared.

So I’m now gonna work out what the other two terms are gonna be. Are they gonna be positive or negative in here? And presumably, they’re going to involve 𝑏. Well I’ve got to generate nine 𝑏 squared. And given that I’ve got an π‘Ž here and an π‘Ž here, the only way I can generate this last term here, nine 𝑏 squared, is if both of these terms purely involve numbers times 𝑏.

So when I do multiply those two terms together I’m gonna 𝑏 times 𝑏, which is 𝑏 squared, and something times something, which gives me nine. So let’s think about what the combinations of those could be. So just sticking into whole numbers for now, one times nine is nine, and three times three is nine.

So if I made those terms one 𝑏 and nine 𝑏 then they multiply together to make nine 𝑏 squared; or if I made them three 𝑏 and three 𝑏, they will also multiply together to make nine 𝑏 squared. But I’ve still got to multiply this term by this term and add it to this term multiplied by this term, so I’ve got combinations of π‘Žs and 𝑏s to consider.

So let’s just do that part of the calculation, so multiplying π‘Ž by some 𝑏s and then adding some 𝑏s multiplied by π‘Ž.

The first of those two terms π‘Ž times something times 𝑏 can be written as something times π‘Žπ‘, and the second one can also be written like that something times π‘Žπ‘ rather than something times 𝑏 times π‘Ž.

And when I add them together, this something and this something need to add together to make negative six; that’s the number of π‘Žπ‘s that I need to generate at the end of this. Well that gives us a problem because so far the options we’ve got are positive one and positive nine or positive three and positive three for those two somethings. Neither of those two pairs added together is gonna generate negative six. So let’s think about what other numbers multiply together to make positive nine.

Well negative one times negative nine makes positive nine, or negative three times negative three makes positive nine. Now which pair of those somethings add together to make negative six? Well it’s negative three and negative three.

So this thing here in front of the 𝑏 is negative three; and this thing here in front of this 𝑏 is also negative three. So let’s tidy up and make a bit of space.

The factored form of that expression is π‘Ž minus three 𝑏 times π‘Ž minus three 𝑏. Well let’s just check our answer. π‘Ž times π‘Ž is π‘Ž squared; π‘Ž times negative three 𝑏 is negative three π‘Žπ‘; negative three 𝑏 times π‘Ž is negative three π‘π‘Ž β€” but it doesn’t matter if we swap those two round, that’s the same as negative three π‘Žπ‘, so I’m going to write that that way round β€” and negative three 𝑏 times negative three 𝑏 is positive nine 𝑏 squared.

Now in the middle, we’ve got negative three π‘Žπ‘ take away another three π‘Žπ‘, so that’s negative six π‘Žπ‘. So when we checked it we do in fact get the same expression that we started off with.

So this is our answer. But you’ll notice that both of the expressions in each of the parenthesis are the same, so we’ve got π‘Ž minus three 𝑏 times π‘Ž minus three 𝑏. That’s π‘Ž minus three 𝑏 all squared, so the most efficient form to write our answer is just that: π‘Ž minus three 𝑏 all squared.