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Factor. If an expression is prime, so indicate.$$-28 u^{3} v^{3}+26 u^{2} v^{4}-6 u v^{5}$$

$-2uv^{3}(2 u-v)(7 u-3 v)$

Algebra

Chapter 6

Factoring and Quadratic Equations

Section 3

Factoring Trinomials of the Form $a x^{2}+b x+c 454$

Equations and Inequalities

Quadratic Functions

Campbell University

Baylor University

Idaho State University

Lectures

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Okay, so let's start by rearranging on terms in terms of highest powers. So we have negative six u Actually, we'll leave it like that. You can factor out, eh? Negative too. As well as a U and a be cute. So we get a good 28. Throughout. Have been able to is 14. You took out one you. So we have squared remaining and no more peas who are 26. Divide that by negative too. You get negative. 13. We took one you away and want to be four and three V's. So we have one be remaining. And then there were six divided by two. Negative too. You get positive. Three the U S taking away. And three bees. So we have to These were reading. Okay. No, that were you squared. Interview squid. All right, binomial, we'll have you in France and a V at the end for both terms. Okay, so we have 14 so we can either make it 14 times one or seven and two. So let's try seven and two are ending. Coefficient has a three. So we'll take a three here and see uncle's. They're both negative since we need the ending coefficient to be positive, but the middle term to be negative. Okay, so let's test if these two terms are correct. So we get 14. You squared minus seven U B minus six, U B minus or a plus three p squared seven plus six leaves me thinking of their team u V, which is what I want so that use the correct characterization.

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