We've seen how to multiply polynomials, so now let's see how we can undo that process. In this section, we'll learn how to divide polynomials and how to check our answer. Here's a polynomial, and we're going to divide it by one monomial, a term. This problem's actually very similar to the problems we've seen in exponents, and in fact, you already have the skills to solve it. We can write this another way by dividing the entire polynomial by 2x. These two forms are equivalent. Next, we divide 2x into each of the terms. So using what you know about exponents and division, I want you to find the answer or the quotient to this division problem.
For the coefficients, we just divide them. We know 8 divided by 2 makes 4. For the bases of x, however, we want to subtract their exponents. We have 7 minus positive 1, which is 6. Another way to think about it is we would have x to the would be our second term. 12 divded by 2 makes 6. And x to the 5th divided by x to the 1 makes x to the 4th. Our last term should be negative 3x squared. 6 divided by 2 equals negative 3. We know it's negative because the sign is negative. And then x to the 3rd divided by x to the 1 is x squared, or x to the right.
What's great about division is we can always check our answer. We know division and multiplication are inverse operations of each other, so if I take my answer were originally divided by, we should get our same polynomial. 2x Times 4x to the 6 is 8x to the 7th, 2x times 6x to the 4th is 12x to the 5th And 2x times negative 3x squared is negative 6x cubed, this is great. We know our polynomials match, so this must be the correct answer to our division problem.
Let's consider the same problem with a constant added to our dividend. The dividend is simply the thing that we're dividing into, and the divisor is the second thing that we're dividing by. How would adding this constant 4 change our answer? What new term would we need to include? Take some time to think about this, and then when you're ready, write your answer in this box.
We divide our polynomial by 2x, which means we really divide each term by 2x. The first three terms would stay the same, because that's the same division we performed earlier. For the last term, however, we have 4 divided by 2, which is positive 2 divided by an x and x is in the denominator. You might have written this as your answer, which is correct as well. A negative exponent moves the base from the numerator to the denominator. And just like before, if we wanted to check, we can multiply this entire polynomial by positive 2x, and we'll get this as our answer.
Let's try something a little more difficult. Here's a polynomial with multiple variables. We have a and b. We're going to divide this polynomial by 5ab squared. We divide each of these terms by 5ab squared. So now, you tell me. Using what you know about division and exponents, what would the answer be?
For this first term, we'll have 5 divided by 5, which equals 1, we'll have a cubed in the numerator and one b in the numerator as well. The second term would be, 8a squared b, we'll have a squared in the numerator, and one b in the numerator as well. For the third term, we'll have positive 3a, we'll have one a left in the numerator, and b squared, divided by b squared, makes 1, so just positive 3a for the third term. And finally, for the last term, we'll just have negative 12, 60 divided by 5 equals 12, and the sign is negative, so we need to make sure we include that here. Ab Squared divided by ab squared would just equal 1, that's why we're only left with negative 12. This would be our answer to our division problem.
Here's another problem for you to try out, this one's tough. Think about what numbers and variables will appear in the numerator and the denominator, take your time and as always good luck.
Just like all the other problems, we're going to start by dividing each term by our divisor, 6fg squared h. 3/6 Reduces to 1/2, so that's going to start my answer. F cubed divided by f would leave me with f squared in the numerator, and then g squared divided by g squared equals 1. And then h 4th divided by h equals h cubed. Our second term would be 4fh squared, 24 divided by 6 equals 4 and f squared divided by f makes f to the 1. The g squared divided by g squared simplifies to 1 again, and then h cubed divided by h is equal to h squared. And finally for our last term, we'll have negative 3 halves, f times h. F squared divided by f is equal to f, and our g squared divided by g squared is equal to leaves us with h to the 1. This first term and this last term, were a little tricky, since we had fractions, but hopefully you got most of the variables right. If you got everything right, excellent work.
But what if we want to divide by something other than a monomial? What if I wanted to do this? Here we're dividing a polynomial by a binomial. You might not know it, but you already have the skills needed to solve this problem. Let's look at how. We're going to compare this to a division problem that you've seen before. We're going to divide 4 into 852, and I want you to fill in the boxes for the division problem. Don't just put the answers here. Complete everything as if you were doing the long division.
We know 4 divides into 8 two times. So we put a 2 above the 8. 2 times 4 is 8, so we subtract 8 from 8 to get 0. To continue our long division problem, we bring down the 5 and then ask, 4 divides into 5 how many times? Well, that's just once. 1 times 4 is 4, and then we can do our subtraction. 5 minus 4 equals into 12 three times. So our final answer is 213.
So why would I have you enter all those numbers for long division? Are we only concerned about the answer? Well, no. Math is about reasoning and we can use that reasoning in new situations like here. We'll use our understanding of dividing numbers to divide polynomials. To divide x plus 5 into x squared plus 7 x plus 10, we're going to start with these first two terms. I want to think about the expression that I could put here to multiply by x to get x squared. In other words, I'm taking x squared and dividing it by x. That variable must be x. Next, we multiply x by x and positive 5, and we get x squared plus 5x . This step was very similar to multiplying 2 times 4 to get the 8. Now that we're here, we want to subtract. So I want to subtract this entire amount or expression. This is really a negative 1 times each of these quantities. So we want to make sure that we change both of the signs. When I subtract the amount I'll have negative x squared and negative 5x. Both of the signs change. X squared minus x squared equals 0, and 7x minus 5x equals 2x. Next we bring down the ten to continue the next step of our long division problem. Now I want you to try that same process again. You're going to divide x plus 5 into 2x plus 10. What number would go here? And then when you multiply this number by x plus 5, what two expressions do you get here? Then you're going to subtract this quantity, and put in your final answers down here.
When we divided x plus 5 into 2x plus 10, we want to look at these first two terms. We want to ask ourselves, what number times x equals 2x? Well, that must be positive 2. 2 times x is 2x and 2 times 5 is 10. You can also get this number by dividing 2x, this first term, by the first term that we're dividing by, x. 2x divided by x, is positive 2. Now we need to subtract this expression. We know subtracting an expression changes both of the signs. We have negative 2x and negative 10. 2x minus 2x equals 0, and positive 10 minus 10 is also equal to 0. So when we divide this polynomial by x plus 5, we get an answer of x plus 2.
Finally, we can check our answer by multiplying x plus 2 times the divisor x plus 5. X times x is x squared and x times positive is positive 5 x. 2 times x is positive x and 2 times 5 is positive 10. And finally, if we combine our like terms, we get x squared plus 7 x plus 10, our original polynomial.
Here's a more challenging problem. For this one, I'm going to draw in the boxes, and I want you to fill them in for completing the long division process. Here are the boxes that you can fill in for the long division. You want to make sure that you put the original product, once you multiply this term, by 3x and negative 2. Also include the appropriate sign in the box. If it's positive, don't put a sign, but if it's negative, do include it.
I want to think about what number multiplied by 3x is 12x squared. That must be term of our divisor, 3x. 12x squared divided by 3x is positive 4x. 4x times 3x is 12x squared. And 4x times negative 2 is negative 8x. Now we're ready to subtract this expression, which means we change both of these signs. The first term is negative and the second term becomes positive. 12x square minus 12 square equals 0, and 1x plus 8x equals 9x. Fantasic work if you got all four of those boxes.
Finish solving this problem by filling in these boxes. You're going to do the division process one more time. Then write your final answer, or your quotient, here in this box.
We know a positive 3 should go here, since 3 times 3x is equal to 9x. Remember, to find this number we can also divide 9x, this first term, by the first term of our divisor, 3x, 9x divided by 3x is positive 3. 3 Times 3x is positive 9x, and which changes both of the signs. This leaves us with 0 as the remainder. So 4x plus 3 is the quotient, our answer, when we divide this polynomial by 3x minus
Here's our first practice problem with division. Divide this polynomial by 4k cubed.
That's an amazing work if you got that correct. We know that 12 divided by 4 equals 3, and k to the 5th divided by k to the 3rd is k squared. 28 divided by 4 equals 7. And we're left with 1k when we compared k to the 4th, to k to the 3rd we'll have one k left in our numerator. Next we'll have negative 2 for negative have any variables here. And finally, we're left with 3 4th as a fraction and then k squared in the denominator. We have more ks down here then in the numerator. At these exponents are giving you trouble, it would be a great idea to review. Try going back to properties of exponents and solving those problems.
Here's the second practice problem. Good luck.
To start this division problem, we want to think about what number, times 3a, is times positive 2, is positive 4a. Next we subtract this second expression. When we do subtraction, we simply change all the signs. We distribute the negative 1. a, then we just bring down our negative 14 to continue long division. Next we want to think about what number times 3a is equal to negative 21a. The number must be negative 7, negative 7 times 3a is negative 21a, and negative 7 times positive 2 is negative 14. When we subtract this expression, we really distribute the negative 1 which changes both of the signs. Adding our like-terms together, we get 0. 2a Minus 7 is our answer and we can check it by, multiplying it by 3a plus 2.
Try this one out.
We divide each term by negative 7x squared. 21 divided by negative 7 equals negative 3, and x squared divided by x squared is equal to 1. So this first term is simply negative 3. For the second one, I have 14 divided by negative 7, which equals negative 2. For the xes, I'll have one more x in the denominator than I will in the numerator. So I have minus a negative 2 divided by x. We're subtracting a negative number here. So really, this is just addition. And finally, 42 divided by negative 7 equals negative 6, and then we're left with x squared and the denominator. This one was pretty tricky, so great work if you got it.
Try dividing 5x minus 4 into this polynomial. Write your answer here.
You're really mastering division, if you got 3x plus 7. First we think about what number times 5x, would give us 15x squared, that must be 3x, 3x times 5x is expression, which changes both of the signs. 15x Squared minus 15x squared is 0, and 23x plus 12x is 35x, we bring down the negative 28 to continue our long division. Repeating our process one more time, we want to think about what number times 5 x gives us 35 x? That should be positive 7. Positive 7 times 5x, is 35x, and positive 7 times negative 4 is negative 28. We subtract this expression, which changes both of the signs, this leaves us with a remainder of