In Exercises 1–16, divide using long division. State the quotient...

Question

In Exercises 1–16, divide using long division. State the quotient, and the remainder, r(x). (x^3+5x^2+7x+2)÷(x+2)

Accepted Answer

Welcome back. I'm so glad you're here. We have got this long division problem. Our dividend is X cubed plus 10 X squared plus 27 X plus and that's being divided by our divisor which is X plus five. All right. Where do we start with? Long division? Well, step one is to divide. Well, okay. More specifically what are we dividing? We're going to take the first term of our dividend. So in this case X cubed and divided by the first term in our divisor. So X X cubed divided by X is X squared. I'm going to write the X squared right above the term in the dividend that has the same degree of X. So X squared right above the 10 X squared. Step two is to multiply What are we going to multiply? We're going to take this new term we have in our quotient and multiply it by all the terms in the divisor. So X squared times X. That is X cubed X squared times five. That's a plus five X squared. There's step two. Step 3 is to subtract what are we going to subtract? We're going to subtract these two new terms in our work here from the two terms right above them with the same degrees in the dividend. So an X cubed minus and X cubed. That's zero. Perfect. That's exactly what we wanted there and and 10 X squared minus five X squared will give us five X squared. Note that I put those two terms in our work in parentheses. So that we subtract both terms and not just the first term written there. Step four is to bring down what are we going to bring down? We bring down the next term from our dividend so that it's in our work and we have the same number of terms in our work as we do in our divisor. So five X squared plus 27 X. That's two terms and the divisor has two terms X plus five. And then we start over. We go back and divide. What do we divide the first term in our work here divided by the first term in the divisor. So we have a five X squared divided by X and five X squared divided by X is plus five X. Now we keep going with multiplication. So we take the new term in our quotient, multiplied by both terms. In our divisor five x times X is five X squared perfect. It matches the one right above it. Five x times five is plus 25 X continuing with subtraction. We're going to subtract both of these terms from the ones above them. In the work five X squared minus five X squared zero. Perfect. 27 X minus 25 X is two X. Step four. Bring down, will bring down this plus 10 and now continuing on with our step one divide, divide the first term in our work here. Two X by the first term. In our divisor by X to X divided by X is to so we write a plus two up here in the quotient and we multiply our new term in the quotient by both terms in the divisor. So we take two times X, which is two X and two times five, which is plus 10. Now, step three subtract both of these terms. Two X minus two. X. Zero. Perfect. That's what we want. 10 minus 10. That is also zero. Alright, when there's nothing left to bring down for step four, then everything we've got here is the remainder and in this case we have a remainder of nothing of zero. So our quotient here, X squared plus five X plus two. That is the answer. We look at our answer choices and this matches with answer choice. C Well done. We'll catch you on the next one.

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In Exercises 1–16, divide using long division. State the quotient, and the remainder, r(x). (x^3+5x^2+7x+2)÷(x+2)

Dividing Polynomials practice set