In Exercises 103–114, factor completely. 6x⁴+35x

Question

In Exercises 103–114, factor completely. 6x⁴+35x²−6

Accepted Answer

hey everyone here we are asked to factor the expression three X to the fourth power plus 13 X squared minus 10. Now here our highest power is four. But we typically factor expressions with the highest power being too. So we can make this expression easier by letting V be equal to X squared. And this allows us to rewrite the given expression as three V squared plus 13 V minus 10. Which is much easier to work with. So we can start factoring Now. We want to begin by looking at this first term here of three V squared and factoring it into our first terms for each set of parentheses. So in the first set we'll have three V. And in the second set we'll have V. Now from here we can move on to finding our second terms. So we want to look at this negative 10 and note that the product of our second terms must be equal to this negative 10. So this allows for many different combinations. So let's start with positive two and negative five. Then negative to positive five, positive five and negative two negative five and positive to positive one and negative 10. Negative one and positive 10 positive 10 and negative one and finally negative 10 and positive one. Now from here we have to determine the appropriate combination. So we have to look at this second term here of 13 V. And note that the some of the products of the inside and outside terms must be equal to this 13 V. So let's test out this first set here of positive two and negative five. Which gives us the quantity of three V plus two times the quantity of vi minus five. So multiplying the outside terms, we have negative 15 V plus the product of the inside terms which is two V. Gives us negative 13 V. Which is not equal to our second term. So we know this combination doesn't work. So we have to move on to the second combination. So the second combination gives us the quantity of three V minus two times the quantity of V plus five repeating the same process. The product of the outside terms which is 15 V. Plus the product of the inside terms negative two V gives us 13 V. Which is our second term. So we know this combination does work. So we know in terms of V our expression factors out into the quantity of three v minus two times the quantity of V plus five. Now all that is left is to bring back in this X squared. So our final answer is the quantity of three X squared minus two times the quantity of X squared plus five which is choice D. Thanks for watching. I hope you found this video helpful and I'll see you again next time

In Exercises 103–114, factor completely. 6x⁴+35x²−6

Key Concepts

Factoring Polynomials

Quadratic Form

Zero Product Property

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In Exercises 103–114, factor completely. 6x4+35x2−6

Key Concepts

Factoring Polynomials

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Zero Product Property

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In Exercises 103–114, factor completely. 6x⁴+35x²−6