Perfect-Square Trinomials

Well, the first term, x², 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:

But what was the original binomial that they'd squared?

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:

The first term, 16x², is the square of 4x, and the last term, 36, is the square of 6.

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:

It's a match to the original quadratic they gave me, so that quadratic fits the pattern of being a perfect square:

I'll plug the 4x and the −6 into the pattern to get the original squared-binomial form:

The first term, 4x², 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:

However, looking back at the original quadratic, it had a middle term of −25x, and this does not match what the pattern requires. So:

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 x⁴, whose square root is x². The third term is 1, whose square root is just 1. Does the middle term, 2x², fit the pattern for perfect-square binomials? I'll check:

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:

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:

Putting the square on everything, I end up with a fully-factoring answer of: