Factor completely, or state that the polynomial is prime. 15x^3+3...

Question

Factor completely, or state that the polynomial is prime. 15x^3+3x^2

Accepted Answer

Hi there. So this problem is the fact that they give them pollinate more completely categorize the pollen with your prime. If it doesn't get fact arised furthermore. So our polynomial are we write this eight X to the fourth plus two X. Now we want to see if we can take some stuff out of this polynomial. Simplify it. Let's take a look. The first thing I have to do is see if I can take anything out of the coefficients. The coefficients of course are the numbers in front of our variables eight and two. Let's see if there's anything common between those two numbers that I can take out. One way I like to do it is by dividing my larger coefficient by my smaller coefficient. 8/2. I would get more. So this is the whole number which means two does go into eight. So we can use to as one of our factors. So go ahead and rewrite this. Taking a two out of everything. Now I do that too is coming out of the 8-4 And we're taking a two From the two x. Which is just going to be taking two of the two, which is just a one. They're really just leaving the X. They're taking the two backyard. Now we want to check and see if we can take out an exponent. So we go back to my exponents. We have X to the fourth and we have X. Someone do the same thing I did with the coefficients. I'll save my largest exponent X to the fourth and divide it by my lowest X funding which is X. Now when we divide exponents, we have to remember that from our book, dividing exponents will subtract the degrees of each exponents. So what I mean by that is we have X to the fourth of our X. We're gonna get X to the ford or subtracting the exponents of now the degree of the bottom as such as X. We know it's actually X the first So this is X to the 4 -1. I'm gonna certify that. Get X to the third so we can actually take out an X from all of our remaining terms. So if I take out that X, I get two X, then I multiplied by what's left. We have a four left. We set out one X from the X to the fourth, receive that as X to the third. Take out an X from the X. That just leaves it as one. So the answer to our factor is two X times four X cubed minus one. Which means our answer is answer to be okay. I want to help solve the problem. Thank you for watching. Goodbye

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