In Exercises 17–32, divide using synthetic division. (x^5+4x^4−3x...

Question

In Exercises 17–32, divide using synthetic division. (x^5+4x^4−3x^2+2x+3)÷(x−3)

Accepted Answer

Hey everyone welcome back in this video. We're going to be using synthetic division to find the quotient of the following polynomial. We have X to the five minus eight X to the four minus four X squared plus five X plus one. All divided by x minus one. When we're doing synthetic division we want to double check that the divisor or the term in the denominator is a degree one polynomial in this case we have x minus one. Okay, as our divisor highest exponent is one. And so that's okay. That's a degree one polynomial. So we can go ahead. Now we're gonna take the numerator X to the five plus eight X to the four minus four X squared plus five X plus one. We're going to rewrite it. And when we do our synthetic division we need to take the coefficient of every single term from highest exponents. All the way down to the constant and that includes terms that are missing. So in this case we're just going to rewrite it to include those missing terms. We have X to the five plus eight X to the four plus zero. Execute minus four X squared plus five X plus one. Alright, so we've rewritten our polynomial. We've checked that we can use the denominator. Now let's draw our synthetic division table and fill in the values. The first value we're going to put in here is to the left of the table and it's going to be a one. Ok. Now, where does that come from? Well it comes from the denominator or the divisor X minus one. What we do is we essentially set x minus one equal to zero and solve for X gives us X equals one. Which is the value we use on the left side of the table. It's like finding the root there. Alright, for the rest of the top row, we're going to fill in with the coefficient corresponding to the terms inner numerator, that polynomial that we wrote out in longer form. Okay. And we're gonna start on the left with the coefficient corresponding to the highest exponent. All the way down on the right to the coefficient corresponding with the constant term. We're going to start with one corresponding to X to the five, eight corresponding to X eight X to the 40 corresponding to the X cubed term minus four corresponding to the X squared term, five corresponding to the linear term and finally one corresponding to our constant term. Okay, now the first step in our synthetic division is to bring this one inside the table down to the bottom and we're just going to rewrite it. Okay now what we're gonna do is we're gonna multiply values from the bottom of our table up to this number on the left. Okay. And so one times one is going to be one and we're just gonna put it right in here to the right and above. Now when we have numbers in a column in our synthetic division we add them. Okay, so eight plus one is going to be nine. Now we have a new number down here and we're gonna add it or multi story we're gonna multiply it to the one on the left, nine times one is nine. And now in the column we add zero plus nine is going to be nine. We have a new number at the bottom. It's going to multiply to the 19 times one. Going to give us nine. We add again in our column -4-plus 9 is going to be five. Multiplying five times the one on the left gives us five. We add in the column five plus five is 10 And one last time 10 times the one on the left hand side. That gives us 10. And adding in the column one plus 10 gives us 11. Alright, so we've done the synthetic division and now we need to interpret the results. So this term on the far right here Is going to be a remainder. So 11 is going to be a remainder. We can call it are now for the other terms. Well let's look at what we started with we started with a polynomial of degree five. The highest exponent was X to the five. And we're dividing it by a polynomial of degree one. Okay. We talked about how that x minus one. In the denominator was a degree one. So what that is going to do it's going to reduce the degree by one. Okay, so we started with X to the five. Now our highest exponent is going to be X to the four. Okay. And these values in our synthetic division table that we found are going to correspond to the coefficients. And again, just like when we wrote the top row, the coefficients correspond from left to right with the exponents highest exponents all the way down to the constant. So we have X to the four starting with And now our next coefficient is nine from our table and we're just going to reduce the exponents. So now we have X cubed. The next coefficient in our table is nine. And this time we have X squared. Okay. Going down a degree each time. Our next coefficient is five. That's going to correspond with the linear term. And finally we have 10 which is going to be our constant. Okay now we need to add in our remainder. So we have plus 11 and this does not get did not get divided nicely. And so it's going to still be divided by the original divisor X minus one. Okay, so this here X to the four plus nine X cubed plus nine X squared plus five. X plus 10 plus 11 over X minus one. That's going to be your answer. And that's going to correspond with answer. Be alright. Thanks everyone for watching. Hope you enjoyed it to you in the next video

In Exercises 17–32, divide using synthetic division. (x^5+4x^4−3x^2+2x+3)÷(x−3)

Key Concepts

Synthetic Division

Polynomial Functions

Remainder Theorem