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## Algebra 2

### Course: Algebra 2 > Unit 3

Lesson 5: Factoring using structure- Identifying quadratic patterns
- Identify quadratic patterns
- Factorization with substitution
- Factorization with substitution
- Factoring using the perfect square pattern
- Factoring using the difference of squares pattern
- Factor polynomials using structure

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# Identifying quadratic patterns

Factoring using structure involves recognizing patterns in expressions, like the difference of squares or perfect square trinomials, and using these patterns to break down complex expressions into simpler ones. As an example, you’ll determine whether the polynomial 9x⁸+6x⁴y+y² can be factored using the perfect square pattern or the difference of squares pattern (or neither).

## Want to join the conversation?

- It seems there has been a glitch in the transition to the new updated Algebra courses. The videos won't play. I'm guessing the path to the source files for the videos weren't updated in the html or something (I've done some website building).(21 votes)
- Didn't understand a single thing. This'll take a while to understand. If there is a different way someone can explain this, it is welcomed. My brain hurts.(15 votes)
- For anyone struggling with this, I found that I needed to have a really strong understanding of the "Factoring Higher Degree Polynomials" topic before understanding this.(4 votes)

- Are these the only quadratic patterns? Are there any others?(10 votes)
- These are not the only ways a quadratic can be factored. For example x^2 + 2x can be factored as x(x + 2).(8 votes)

- So the ones that can not use any patterns be a prime polynomial?

Thank you.(6 votes)- Patterns are for special relationships such as perfect square binomials or difference of perfect squares. There are many more factorable trinomials that do not fit the patterns, so the answer is no. If you have x^2 + 15x + 50, this does not fit the patterns, but there are two numbers which multiply to 50 and add to be 15 (10 and 5), so it is not prime, it is (x+10)(x+5).(6 votes)

- On answers, A and B wouldn't (U-V)^2 be equal to (U+V)(U-V)?

and why is "(U+V)^2 or (U-V)^2" aren't they different?(4 votes)- (U-V)^2 is not equal to (U+V)(U-V). (U+V)(U-V) evaluates to U^2 - V^2, while (U-V)^2 evaluates to U^2 - 2UV + V^2.

For the second part of your question, yes (U+V)^2 and (U-V)^2 are different things but the question is simply looking for similar patterns so they group the two together.(3 votes)

- Hi there,

One of the practice quiz questions regarding factoring with substitution had the following question:

81x^6-(y+1)^2

It then asked to further express as (U+V)(U-V)

My answer was (9x^3+y+1)(9x^3-y+1); however, the correct answer was (9x^3+y+1)(9x^3-y-1). Why would V (y+1) turn into (y-1) when I'm following the pattern of U-V?(3 votes)- Think about this, you want to minus (1+3) this whole thing from 9, would you write 9-1+3 ? No! If you write that you would get the result 11, but actually 9 minus the whole thing (1+3) is 5. Why? Because you are going to minus
**the whole thing**, you would need to either:

1. add parenthesis around 1+3 first, so you can calculate that first and then minus the result from 9

or

2. change the + sign to -, you would get 9-1-3, now you minus them separately

Now the problem is fixed, when you minus the sum of two numbers from another number, you would either calculate the sum first and then minus, or minus them each separately, not to minus one of them and then add the other.(1 vote)

- For the
**Identifying Quadratic Patterns**, I watched the video, did the practice, I just can't get to proficient. I am massively confused with the topic. Can anyone help me?(3 votes)- These patterns are common ways to factor quadratics. (U+V)^2 or (U-V)^2 are the factorizations of perfect square trinomials. You use them anytime the expression is in the pattern U^2+2UV+V^2 or U^2-2UV+V^2.

For example: x^2+2x+1 uses the (U+V)^2 pattern because it factors into (x+1)^2, where U=x and V=1.

(U+V)(U-V) is always used when the expression is in the pattern U^2-V^2 because that's what it expands into.

Ex: 9x^2-16y^2 uses the (U+V)(U-V) pattern since it factors into (3x+4y)(3x-4y), where U=3x and V=4y.

Hope this clears things up a little bit! and just as a note, a and b are often used instead of U and V.(3 votes)

- At1:40, Sal mentions the whole 2UV thing. He includes x and y in 2UV. In the next example, he says at2:45that because there is a y, (U+V)^2 or (U-V)^2 is incorrect. What am I to do?(2 votes)
- Realizing that in the second example that both squared binomial rules do not apply, you should put that you cannot use either of such formulas.

If you also are unsure on why you cannot use the formula,

Note that as a perfect squared binomial, V=5.

If V=5, then the product should be 4x^6+10x^3+25.**However**! You will notice that there is an extra y in the 2UV term. Since V=5 and*not*= 5y, then you will realize that the polynomial in the problem is exempted from those formulas.

Hopefully that helps !(3 votes)

- This is super weird and confusing.(3 votes)

## Video transcript

- [Instructor] We're
told that we wanna factor the following expression and they ask us which pattern can we use
to factor the expression? And U and V are either constant integers or single-variable expressions. So we'll do this one together and then we'll have a few more examples where I'll encourage you to pause the video. So when they're talking about patterns they're really saying, "Hey, can we say "that some of these can generally form "a pattern that matches what we have here "and then we can use
that pattern to factor "it into one of these forms?" What do I mean by that? Well, let's just imagine something like U plus V squared. We've squared binomials in the past. This is going to be equal to U squared plus two times the product of these terms so two U V and then plus V squared. Now, when you look at
this polynomial right over here, it actually has this form if you look at it carefully. How can it have this form? Well, if we view U squared as nine X to the eighth then that means that U is, and let me write it as a capital U, U is equal to three X to the fourth. 'Cause notice if you square this you're gonna get nine X to the eighth. So this right over here is U squared. And if we said that V squared is equal to Y squared, so if this
is capital V squared then that means that V is equal to Y. And then this would have
to be two times U V. Is it? Well, see if I multiply
U times V I get three X to the fourth Y and then two times that is indeed six X to the fourth Y. So this right over here is two U V. So notice this polynomial,
this higher-degree polynomial can be
expressed in this pattern, which means it can be factored this way. So when they say which pattern can we use to factor this expression,
well, I would use a pattern for U plus V squared. So I would go with that
choice right over there. Let's do a few more examples. So here once again we're
told the same thing. We're given a different expression and they're asking us
what pattern can we use to factor the expression? So I have these three terms here. It looks like maybe I could use, I can see a perfect square here. Let's see if that works. If this is U squared, if this is U squared then that means that
U is going to be equal to two X to the third power. And if this is V squared then that means that V is equal to five. Now is this equal to two times U V? Well, let's see, two times
U V would be equal to, well, you're not gonna have any Y in it so this is not going to be two U V. So this actually is not fitting
the perfect square pattern. So we could rule this out. And both of these are
perfect squares of some form. One just has a, I guess
you'd say adding V. The other one is subtracting V. This right over here,
if I were to multiply this out, this is going to be equal to, this is a difference of squares
and we've seen this before. This is U squared minus V squared. So you wouldn't have a
three-term polynomial like that. So we could rule that one out. So I would pick that we can't
use any of the patterns. Let's do yet another example. And I encourage you,
pause the video and see if you can work this one out on your own. So the same idea, they wanna factor the following expression. And this one essentially has two terms. We have a term here and
we have a term here. They both look like they are the square of something and we have
a difference of squares. So this is making me feel pretty good about this pattern but
let's see if that works out. Remember, U plus V
times U minus V is equal to U squared minus V squared. So if this is equal to U
squared then that means that capital U is equal to six X squared. That works. And if this is equal to V squared, well, that means that V is
equal to Y plus three. So this is fitting this
pattern right over here. And they're just asking
us to say what pattern can we use to factor the expression. They're not asking us
to actually factor it. So we'll just pick this choice. But once you identify
the pattern it's actually pretty straightforward to factor it because if you say this
is just going to factor into U plus V times U
minus V, well, U plus V is going to be six X squared plus V plus Y plus three times U minus V. U is six X squared. Minus V is minus, we could write minus Y plus three or we could
distribute the negative sign. But either way this might
make it a little bit clearer what we just did. We used a pattern to
factor this higher-degree polynomial which is essentially just a difference of squares. Let's do one last example. So here once again we
wanna factor an expression. Which pattern can we use? Pause the video. All right, so we have two terms here. So it looks like it might be a difference of squares if we set U is equal to seven and then this would be U U squared. But then what can we square to get 10 X to the third power? Remember, we wanna have
integer exponents here. And the square root of
10 X to the third power, if I were to take the square root of 10 X to the third power it would be something a little bit involved
like the square root of 10 times X times the square root of X to the third power and
I'm not going to get an integer exponent here. So it doesn't look like I can
express this as V squared. So I would go with that we
can't use any of the patterns. And we're done.