Solve: x^4−6x^3+4x^2+15x+4=0.

Question

Accepted Answer

Welcome back. I am so glad you're here. We are given this equation A to the fourth plus A. To the third minus seven. A squared minus five A plus 10 equals zero. And we're asked to solve. Alright well how do we solve for a The degree here is four. Well we recall from previous lessons that we can utilize the rational zeros theorem to find out what possible rational zeros there are for this equation. So the rational zeros theorem, excuse me, has us identify the constant term. That here is 10, that's R. P. And it has this identify the leading coefficient and that here is one. That's our cue. And we're going to take all of the factors of P and divide them by the factors of Q. And that would be all of our possible rational zeros. All of our possible solutions. And so the rational ones anyway. And so what are our factors of 10? Well, tens factors are plus or minus one plus or minus two Plus or -5 and plus or -10. How about for Q? Well the possible factors for Q are all plus or minus one. So dividing everything here by plus or minus one is just going to give us plus or minus one plus or minus two plus or minus five and plus or minus 10. So we've got eight possible solutions here for our rational zeros. And now what do we do with those while we can utilize synthetic division to identify which ones are actual zeros. So I'm just going to start with one because it's at the beginning. and so now recalling our previous lessons on synthetic division, I'm going to bring down the coefficients in the constant term. So we've got 11 negative seven, negative five and 10. And now our steps for synthetic division start off bringing down the one and now we'll multiply we've got one times one is one back to adding one plus, one is two back to multiplying one times two is two adding negative seven plus two is negative five. Multiplying one times negative five is negative five adding negative five plus negative five is negative 10. Multiplying one times negative 10 is negative tin. And adding 10 plus negative 10 is zero. And look at that one is a solution here, it is a rational zero. And so it is one solution for a. And what's our new equation Now, will we recall that our answers here are going to be our new coefficients and constant term for an equation that is one degree less than the one we started with. So the degree of our starting equation was four. So now this one is going to be the leading coefficient for a. Two. The third one degree less than a to the fourth. So we've got eight of the third plus two, A squared minus five, A minus 10 equals zero. And again we can evaluate for P and Q. Well negative 10 is P and Q. Is one again. So it's actually going to be those same. eight possible rational zeros plus or minus one plus or minus two. Plus or minus five and plus or minus 10. So there's still eight possibilities to go with for synthetic division here to test out. And at this point I would look since we have some different answer choices here, I would look and see what possibilities line up with A. And for answer choice is A. And B. They have a one and so they're both paired with negative two. So let's test out negative two and ensure that that is correct. So we'll use negative two as our next possible rational zero here and we'll bring down our coefficients and constant terms. So we've got 12 negative five negative 10. And we'll do the synthetic division to ensure that this is a solution. So bring down the one negative two times one is negative two, two plus negative two is zero negative two times zero is zero, negative five plus zero is negative five -2 times negative five is a positive 10 and negative 10 plus 10 is zero. So yes negative two does in fact work. And what's our new formula here? So this one is the leading coefficient of in a with one degree less than our previous equation. So this will be one a squared one degree less than R. A cubed formula. So we've got a squared and there are 08. And then it's just a squared minus five equals zero. And we can solve for a here a squared minus five equals zero. Let's add five to both sides. And we've got a squared equals five. We can take the square root of both sides and we have a equals plus or minus the square root of five. So there are other two solutions for A the negative square +25 and the positive square to five. We look at our answer choices and this matches with answer choice B. Well done, we'll catch you on the next one.

Solve: x^4−6x^3+4x^2+15x+4=0.

Key Concepts

Polynomial Functions

Factoring Polynomials

The Rational Root Theorem