Solve: x^4+2x^3−x^2−4x−2=0

Question

Accepted Answer

Hello today we're going to be solving for X for this higher order polynomial. So the polynomial given to us is X to the fourth plus eight X cubed plus 11, X squared minus 40 X minus 80 is equal to zero. Now, in order to solve for X, we need to go ahead and use the method of possible zeros or as some books call it the PQ method. So P is going to be the constant at the end of your polynomial and Q is going to be the leading coefficient. And so we want to go ahead and list out all the possible zeros for this polynomial. And so that's why we use the factors P over Q. Or in other words we're gonna use the factors of 80 over the factors of one and they could all potentially be plus or minus. So let's go ahead and list out all the possible factors of 80 while 80 has the factors of +1245, 8 10 2040 and 80. And now we want to go ahead and list all the possible possible factors for one, which in this case is just gonna be one. So in order to simplify this, you divide each of these factors by the factors in the denominator. And so all the possible zeros are going to be plus or minus +12458, 10 16 2040 and 80. So what we're gonna do is go ahead and pick one of these factors and synthetically divide our polynomial by that factor. That factor needs to produce a value of zero at the end of the synthetic division in order for it to be a zero while let's go ahead and pick the value negative four. So we're going to synthetically divide our original polynomial by negative four. And that means we're going to list out all the coefficients from our polynomial. So the coefficients are gonna be 18, 11, negative 40 and negative 80. And in order to synthetically divide this, we're going to bring down the first coefficient by itself, which is one. Now what we're gonna put underneath eight is going to be the product of one and negative four. So one times negative four is going to give us negative four and now we're gonna go ahead and add these values together. Eight minus four is going to give us positive four. And we're going to repeat this process until we reach the very end. So four times negative four is going to give us negative and 11 minus 16 is going to give us negative five. Negative five times negative four is going to give us positive 20 and negative 40 plus 20 is going to give us negative 20. Now negative 20 times negative four is going to give us positive and negative 80 plus positive 80 is going to give us zero. So since negative for produced a zero as the remainder for the synthetic division. This is a possible value of X. Or this is a zero of X. So one of the zeros for X is X is equal to negative four. And now we want to go ahead and list out the rest of our polynomial. The way we do this is we list out a new polynomial but with one power less than the original. And we're going to use the new coefficients. So one is gonna be our new leading coefficient. And instead of having X to the fourth, we're going to have X cubed and that's gonna be plus four, X squared minus five, X minus 20. And now we need to go ahead and factor this version. Well we can go ahead and factor this by using factoring by grouping. So we're gonna go ahead and group up the first two terms and the last two terms we can factor out an X squared from the first group. And that's going to give us X squared times X plus four. And on the right hand group we have negative five X and negative 20. We can factor out a negative five. And that's going to give us negative five times the quantity X plus four. And that's going to be equal to zero. Now we go ahead and combine these factors together and that's going to give us X squared minus five times the quantity X plus four is equal to zero. And using the zero property. We can set both of these factors equal to zero and sulfur X. So we get X squared minus five is equal to zero and X plus four is equal to zero. Now on the right hand side we need to go ahead and attract four to both sides of the equation. And that means that X is going to equal to negative four. And that's another value of X. On the left hand side we add five to both sides of the equation and that gives us X square is equal to five. Now in order to get X by itself, we need to take the square root of both sides of this equation and that's going to give us X is equal to the square root of five. Now you need to be careful here because whenever you take a square root it produces a positive and a negative value. So what we have is X is equal to positive square root of five and X is equal to negative square root of five. So there are four parts, there are four values of X. In this case X is equal to negative four comma negative four comma square root of five, comma negative square root of five. Now if you notice we have two repeating values of negative four, we only need to include negative four once since there's two instances of it. So we can go ahead and get rid of one of them. So there are three values for X. X is equal to negative four square root of five and negative square root of five. And that's going to be the answer to this equation. And that means the answer to this problem is going to be a. So I hope this video helps you understanding how to solve for higher order polynomial. And I'll go ahead and see you all in the next video.

Solve: x^4+2x^3−x^2−4x−2=0

Key Concepts

Polynomial Functions

Factoring Polynomials

The Rational Root Theorem