In Exercises 17–24, a) List all possible rational roots. b) Li...

Question

In Exercises 17–24, a) List all possible rational roots. b) List all possible rational roots. c) Use the quotient from part (b) to find the remaining roots and solve the equation. x⁴−2x³−5x²+8x+4=0

Accepted Answer

hey everyone here we are asked to list the possible rational zeros and find an actual zero of the equation. Nine x cubed plus 27 x squared minus 40 X minus 16 is equal to zero. And now from here to start finding the possible rational zeros. We first need to find the factors of the constant and we also need to find the factors of the leading coefficient. So starting with the constant, which is negative 16. Well 16 has factors of 1248 and 16. And each of these terms get a plus or minus in front of it because each factor can either be positive or negative. And so from here we can start finding the factors of our leading coefficient which is nine. So nine has factors of 13 and nine. And again each of these terms get a plus or minus in front of them. And now from here to find the possible rational zeros. We need to recall the rational zeros theorem which tells us that we can take the factors of the constant and then divide them by the factors of the co the leading coefficient. And so doing this, we can start by dividing each constant factor by one. So we have one divided by one which is plus or minus +12 divided by one which is plus or minus +24 divided by one, which is plus or minus four, then plus or minus eight. And plus or minus 16. And now from here we can move on to dividing each constant factor by three. So we have one divided by three giving us plus or minus one third, two divided by three giving us plus or minus two thirds. We then have four divided by three which is plus or minus four thirds. We then have plus or minus eight thirds and plus or minus 16 3rd. And now from here we can start dividing each factor by nine. So we have one divided by nine which is plus or minus 1/9. Then we have two divided by nine giving us plus or minus two nights. Then four divided by nine giving us plus or minus four nights. We then have plus or minus 8/9 and finally plus or minus 16 nights. And now after we list all of the possible rational zeros. We can start to try to find the actual zero Using synthetic division. So setting up this problem using synthetic division. On the right hand side we have the coefficients of from the equation. So we have 9, 27, negative 40 and negative 16. And on the left hand side we choose one of the possible rational zeros. For simplicity I'm just going to choose positive one. That's a good starting point. So now we can start solving as we normally would with synthetic division. So we first bring down the first coefficient which is nine without making any changes. We then bring the nine up and multiply it by our left side turn which is one. So then we have nine and nine plus 27 gives us 36. We then bring the 36 up and multiply it by one. Giving us 36. Then we add 36 negative 40 giving us negative four. Then we take the negative four up and multiply it by one giving us negative four. And we added to negative 16 giving us negative 20. And now we see that we have a remainder of negative 20 when we use the possible rational zero of one. So we know that one is not a zero because we need the remainder to be zero. So now we can just move on and test another one of our terms from the list we generated. So again setting this up we have the coefficients 27 negative 40 and negative 16 on the right hand side. And on the left hand side This time I'm going to choose negative four. So if we bring down the first term we just have nine by itself and then we bring nine up and multiply it by negative four. This gives us negative 36 negative 36 plus 27 gives us negative nine. Now repeating. We have negative nine times negative four. Giving us positive 36 36 plus negative 40. Gives us negative four. Then bringing negative four up. And multiplying it by negative four. We get positive 16 16 plus negative 16 gives us zero. So we see we have a remainder of zero and therefore negative four is one of our zeros. And so if we look at our giving equation here of nine X cubed plus 27 X squared minus 40 x minus 16. We can now factor out this X plus four and we see that we can do this and then we get the factored equation of nine X squared minus nine x minus four. And we set it equal to zero just how it is given to us. And so now we can see again here that negative four is one of our zeros. And now we can look at this equation here and we have nine X squared minus nine X minus four is equal to zero. And if we use the quadratic formula we see that our additional actual zeros are. So here we would have access equal to four thirds and we also have negative one third. So in summary we can see that our actual zeros are X is equal to negative four negative one third and four thirds. So these three terms are again our actual zeros. And the long list we generated earlier is the list of our possible rational zeros. And that leaves us with answer choice D Thanks for watching. I hope you found this video helpful. And I'll see you again next time

In Exercises 17–24, a) List all possible rational roots. b) List all possible rational roots. c) Use the quotient from part (b) to find the remaining roots and solve the equation. x⁴−2x³−5x²+8x+4=0

Key Concepts

Rational Root Theorem

Polynomial Division (Synthetic or Long Division)

Solving Polynomial Equations by Factoring

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In Exercises 17–24, a) List all possible rational roots. b) List all possible rational roots. c) Use the quotient from part (b) to find the remaining roots and solve the equation. x4−2x3−5x2+8x+4=0

Key Concepts

Rational Root Theorem

Polynomial Division (Synthetic or Long Division)

Solving Polynomial Equations by Factoring

Watch next

In Exercises 17–24, a) List all possible rational roots. b) List all possible rational roots. c) Use the quotient from part (b) to find the remaining roots and solve the equation. x⁴−2x³−5x²+8x+4=0