There is one "special" factoring type that can actually be done using the usual methods for factoring, but, for whatever reason, many texts and instructors make a big deal of treating this case separately. "Perfect square trinomials" are quadratics which are the results of squaring binomials. (Remember that "trinomial" means "three-term polynomial".) For instance:
(x + 3)^{2}
= (x + 3)(x + 3)
= x^{2} + 6x + 9
...so x^{2} + 6x + 9 is a perfect square trinomial.
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Recognizing the pattern to perfect squares isn't a make-or-break issue — these are quadratics that you can factor in the usual way — but noticing the pattern can be a time-saver occasionally, which can be helpful on timed tests.
The trick to seeing this pattern is really quite simple: If the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2, and then compare your result with the original quadratic's middle term. If you've got a match (ignoring the sign), then you've got a perfect-square trinomial. And the original binomial that they'd squared was the sum (or difference) of the square roots of the first and third terms, together with the sign that was on the middle term of the trinomial.
Perfect-square trinomials are of the form:
a^{2}x^{2} ± 2axb + b^{2}
...and are expressed in squared-binomial form as:
(ax ± b)^{2}
How does this look, in practice?
Well, the first term, x^{2}, is the square of x. The third term, 25, is the square of 5. Multiplying these two, I get 5x.
Multiplying this expression by 2, I get 10x. This is what I'm needing to match, in order for the quadratic to fit the pattern of a perfect-square trinomial. Looking at the original quadratic they gave me, I see that the middle term is 10x, which is what I needed. So this is indeed a perfect-square trinomial:
(x)^{2} + 2(x)(5) + (5)^{2}
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But what was the original binomial that they'd squared?
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I know that the first term in the original binomial will be the first square root I found, which was x. The second term will be the second square root I found, which was 5. Looking back at the original quadratic, I see that the sign on the middle term was a "plus". This means that I'll have a "plus" sign between the x and the 5. Then this quadratic is:
a perfect square, with
x^{2} + 10x + 25 = (x + 5)^{2}
The first term, 16x^{2}, is the square of 4x, and the last term, 36, is the square of 6.
(4x)^{2} – 48x + 6^{2}
Actually, since the middle term has a "minus" sign, the 36 will need to be the square of –6 if the pattern is going to work. Just to be sure, I'll make sure that the middle term matches the pattern:
(4x)(–6)(2) = –48x
It's a match to the original quadratic they gave me, so that quadratic fits the pattern of being a perfect square:
(4x)^{2} + (2)(4x)(–6) + (–6)^{2}
I'll plug the 4x and the –6 into the pattern to get the original squared-binomial form:
16x^{2} – 48x + 36 = (4x – 6)^{2}
The first term, 4x^{2}, is the square of 2x, and the last term, 36, is the square of 6 (or, in this case, –6, if this is a perfect square).
According to the pattern for perfect-square trinomials, the middle term must be:
(2x)(–6)(2) = –24x
However, looking back at the original quadratic, it had a middle term of –25x, and this does not match what the pattern requires. So:
this is not a perfect square trinomial.
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If I use the regular methods for factoring quadratic-type polynomials, I can factor this just fine. But what if this is in the homework for the section in my textbook on perfect-square binomials? Naturally, I'm going to be thinking that the author is expecting me to notice a perfect square. So:
The first term is x^{4}, whose square root is x^{2}. The third term is 1, whose square root is just 1. Does the middle term, 2x^{2}, fit the pattern for perfect-square binomials? I'll check:
2(x^{2})(1) = 2x^{2}
It's a match to the original polynomial, so this is a perfect-square trinomial. With the "minus" on the middle term of what they gave me, the original squared-binomial form looks like:
(x^{2} – 1)^{2}
Hmm... The instructions say to "factor fully". That's often a clue that there may be some more factoring that I could, after the usual bit is completed. Can I factor any more here?
Yes, I can. Looking inside the parentheses, I notice that I have a difference of squares, which I can factor:
x^{2} – 1 = (x – 1)(x + 1)
Putting the square on everything, I end up with a fully-factoring answer of:
x^{4} – 2x^{2} + 1 = (x^{2} – 1)^{2}
= ((x – 1)(x + 1))^{2}
= (x – 1)^{2}(x + 1)^{2}
That's really all there is to perfect squares.
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