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### Course: Get ready for Precalculus>Unit 2

Lesson 5: Factoring using structure

Determining 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).
• 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.
• 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.
• Are these the only quadratic patterns? Are there any others?
• These are not the only ways a quadratic can be factored. For example x^2 + 2x can be factored as x(x + 2).
• So the ones that can not use any patterns be a prime polynomial?

Thank you.
• 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).
• 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?
• (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.
• 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?
• 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?
• 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.
• At , Sal mentions the whole 2UV thing. He includes x and y in 2UV. In the next example, he says at that because there is a y, (U+V)^2 or (U-V)^2 is incorrect. What am I to do?
• 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 !
• ok i came up to this question and couldn't answer it as usual

1+ 2+ 3+ ………. 12 = …...........

choices
A. 65 B.70 C.78 D.80

so yeah
(1 vote)
• So this question is asking for the sum of the integers 1 through 12.
1 and 12 make 13
2 and 11 make 13
3 and 10 make 13
and so on until
6 and 7 make 13
This makes 6 pairs of numbers that add to 13, so the answer is
6 * 13 = 78 = C