9/2 Synthetic Division

All right, y'all synthetic division so yesterday we started doing long division. And when we were doing long division, remember that we have to make sure that we have every single term. Within standard form, and so this one is missing a zero X cubed. So we're going to put in our next cube, so we're going to put in a zero X cubed. And then we have a minus ten X squared. Minus two X -14, and we are going to divide by X minus three. Okay, so the way we did this, we said, what we look at the first terms, X, X to the fourth, and we say, what do we multiply times X to get X to the fourth? Or you can say, what is X to the fourth divided by X? That is X cubed. You are going to put the X cubed up top and you're going to distribute the X cubed through the X minus three. And that's going to give you X to the fourth. Minus three X cubed. So you're going to record the X to the fourth minus three X cubed underneath your first two terms of your dividend and then we're going to draw the line change both signs and combined because we're subtracting. So that's going to give us three X cubed. And then we bring down the next term minus ten X squared. And we take a look at it again and you want to find out what do you multiply times X to get three X squared. That is a three X cubed. Sorry, that's three X squared. Or you can ask yourself what is three X cubed divided by X, which is three X squared. And then you're going to multiply that through the whole divisor. So three X squared times X is three X cubed. And three X squared times negative three is negative 9 X squared. Then we're subtracting, so draw the line. Change both sides combined to have negative X squared. We're going to bring down your next term. Then we do the same thing. We look at only the first terms. What do we multiply by X to get negative X squared? That is negative X so we're going to multiply negative X through to X minus three. And that will give us negative X squared. Plus three X and we're going to draw the line and change both sides combined. It's going to give us negative 5 X and we're going to bring down the next term, negative 14. And then we are looking for the number that we multiply times X to get negative 5 X that answer is negative 5. We're going to distribute the negative 5 through the X minus three. And that's going to give us negative 5 X plus 15. Travel line change both sides combined. That gives us negative 29, which is our remainder, so the negative becomes the sign in the polynomial, and we put 29 over what we're dividing by the X minus three, and that is our quotient. So starting today, we're going to do this thing called synthetic division. And the first thing you have to do with synthetic division is find this divisor value, the way you find the divisor value. Is you are going to take X minus three and set it equal to zero. So you take your divisor, set it equal to zero. And solve. So the way we solve this, we add three to both sides that gives us X equals three, our divisor value is three. Then you are going to put all of the coefficients. The same coefficients that you used when we did long division because it needs to be in standard form. And you need to put a zero for anything that's missing. So our coefficient or our number in front of X to the fourth is one. The number in front of X squared is negative ten. I'm sorry. It is not the number and from the X squared is negative ten, but we're missing an X cubed. So we need a zero there. And then our number in front of X is negative two, and our constant is a negative 14. So we just have written down the numbers that are in front, including their signs. Okay? Then what you're going to do is you're going to bring the first number down. So the one is going to go down both below the addition line. And now we're going to multiply. So it says a times K, we're going to multiply one times three, we're going to record that number underneath the next coefficient, and then add zero plus three is three. Then you're going to take that number that's below what I call the addition line. Because it's helping us figure out what we're adding. We're going to multiply it times the divisor value. So three times three is 9. And then we're going to add to the next coefficient. So negative ten plus 9 is negative one. And then we're going to multiply negative one times three. That's negative three right up underneath the next coefficient. Add negative two plus negative three is negative 5. Then you're going to take your negative 5 and multiply times three. That's going to give you negative 15 recorded under the negative 14 and then add them together and get negative 29. So what I want you to notice is the relationship between the coefficients of our quotient and the numbers that we just got and you should notice that they're the same. Okay, so this number down here becomes your answer. The last number is always the remainder. So you put it over the divisor, put it over the X minus three. The number next to it is the coefficient. And then you're going to do X, X squared, X to the third, so X is going to start increasing each time by one. And you're going to put pluses to make it a polynomial, okay? So the answer is X cubed plus three X squared. Minus X -5 -29 over X minus three. Okay. So this looks really complex because it's a complex algorithm. But really it's not that hard, okay? So you find out what your divisor value is, set that equal to zero solve at number three. And then we put the coefficients so the coefficients are two, negative 5, negative 8, 11, and 16. Put up our addition line. We always bring the first number down, because whatever number is underneath, gets multiplied to the divisor value. We don't have a number down there to start off with. So we bring it down. So we're going to take two times three, record that number underneath the negative 5, add the get one. We're going to take one times three, that's three to negative 8. Gives us negative 5. Negative 5 times three is negative 15. Add that to 11. That gives us negative four. Negative four times three is negative 12. And that's 16. That gives us four. Again, the last number is the remainder, you're always going to put it over what you're dividing by. And then put a sign there. The next number is your constant, then you're going to throw an X after the next number, X squared. Put a plus sign, and then X cubed. So your answer is two X cubed. Plus X squared -5 X minus four. Plus four over X minus three. Okay, this last example. Our divisor value is, it's always opposite of what's in there. So it's a negative four. We put our coefficients one, negative three, negative 7, and 6. We bring down that first value one, one times negative four is negative four. Negative four plus three is negative 7. We multiply that negative 7 times negative four. That gives us positive 28. At the 28 plus negative 7, that's 21. We're going to multiply the 24 times negative four. That's negative 84 at the negative 84 plus the 6 to get negative 70. And then this last term is your remainder, put it over X plus four. Then you're constant. X and X squared. So your answer is X squared -7 X plus 21. Minus 78. Over X plus four. Synthetic division only works when the coefficient of your divisor, the coefficient of X in your divisor is one. Otherwise, you have to do long division. All right, the recording I did in class unfortunately did not record this part of the note. So here is my fix for that. And I hope it helps. Let me know if you have any questions.