In Exercises 27–29, divide using long division. (4x^4 +6x^3 + 3x ...

Question

In Exercises 27–29, divide using long division. (4x^4 +6x^3 + 3x - 1) ÷ (2x^2 + 1)

Accepted Answer

Hello today we're going to simplify a polynomial using long division. So the polynomial given to us is seven X to the fourth power minus 21 X cubed plus six X squared minus X plus three. And all of that is going to be divided by X squared plus four. Now we need to be careful with our setup because X squared plus four is missing an X term so we have an X term that is missing. Well how do we include that? X term into X squared plus four. Well in order to do that all we're gonna do is add zero X into this mono meal. That will give us X squared plus zero X plus four. And that's what we're going to divide this polynomial by. Okay so now that we have our setup we can go ahead and start the process of long division. So the easiest way to approach this. We want to go ahead and multiply this Trina meal so that the first term X squared can closely line up with seven X. To the fourth power. Well what's the number that I can multiply X squared by that will give me seven X to the fourth X squared times seven X squared is going to give me seven X to the fourth. So if I multiply this try no meal by seven X squared. What that leaves me with is seven X to the fourth plus zero X cubed Plus 28 x squared. Now keep in mind that the rules of division state that we are subtracting this value so we need to go ahead and distribute this negative into every term here. Doing so is just going to change the signs of our terms. And what we are left with is negative seven X. To the fourth minus zero X cubed minus 28 X squared. And now we can simplify seven X to the fourth minus seven X. To the fourth cancels each other out negative 21 X cubed minus zero X cubed gives me negative X cubed and six X squared minus 28 X squared is going to give me negative 22 X squared. And what I'm gonna do now is bring down my next term which is negative X. And we're going to repeat this process. So what is a number that I can multiply this. Try no meal by so that X squared can closely line up with negative 21 X cubed. Well if I multiply X squared by negative 21 X. What that gives me as negative 21 X cubed. So I'm gonna multiply this trin Amiel by negative X. And what that gives me is negative 21 X cubed minus zero X squared -84 x. And once again we are subtracting this value so we're going to have to distribute the negative into this trin Amiel. Doing so just changes the signs of the trine Amiel. And what I'm left with is positive 21 X cubed plus zero X squared plus 84 X. Now I can simplify -21 x cubed plus 21 x cubed cancels each other out negative 22 X squared plus zero X squared. Leaves me with negative 22 X squared and negative X plus 84 X. Gives me positive 83 X. And I'm gonna go ahead and bring down our next term which is positive three. And we're going to repeat the process. What is a number that I can multiply X squared by to give me negative 22 X squared. Well X squared times negative 22 is going to give me negative 22 X square. So I'm just gonna go ahead and multiply this trin Amiel by negative 22. And what that gives me is negative 22 X squared minus zero X. And negative 22 times four is going to give me negative 88. And once again this values being subtracted. So we're gonna go ahead and distribute this negative sign. Doing so just changes the signs of all the terms. And what we are left with is positive 22 X squared plus zero X plus 88. And now we simplify negative 22 X squared plus X squared cancels each other out 83 X plus zero X gives me 83 X. And 88 plus three is going to give me positive 91. Now at this point right here we cannot multiply X squared by any value that will give us a term that has X to the power of one. So it is not possible to multiply X squared into 83 X. Therefore this value here is going to be our remainder. So now we can go ahead and put together our solution, our solution is going to be our quotient plus the remainder. So the question that we got from the long division is seven X squared minus 21 X minus and 83 X plus 91 is a positive value. So we're going to add our remainder to our quotient and this remainder is still being divided by our original mono meal, X squared plus four. So the way that we write the remainder is going to be the quantity 83 X plus and that's going to be divided by X squared plus four. And this entire thing is going to be the solution to the long division. This also means that the solution to this problem is going to be D. So I hope this video helps you in understanding how to use long division and I'll go ahead and see you all in the next video

Struggling with College Algebra?

In Exercises 27–29, divide using long division. (4x^4 +6x^3 + 3x - 1) ÷ (2x^2 + 1)

Dividing Polynomials practice set