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Synthetic Division & Long Division of Polynomials

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Today, we're going to talk about dividing polynomials using synthetic division and long division. We'll start by doing an example using long division, which you've used before when you're dividing expressions with just numbers. And then we'll do a couple of examples using a method called synthetic division.

For my first example, I'm going to use long division to divide x squared minus 2x minus 15 by x minus 5. I'm going to start by working backwards and thinking of multiplication. What would I multiply by x to give me x squared? That would be another x. So I'll write that x on the top, and then I'll take that x and multiply it by both terms of my divisor. Then I'm going to write that underneath my original dividend. So x times x will give me x squared, and x times negative 5 will give me negative 5x.

I'm going to subtract those two terms from the dividend from the expression above it, the dividend. And that's going to give me-- x squared minus x squared will give me zero, or it just cancels out. And then negative 2x minus negative 5x is going to give me a positive 3x. And I'm also going to bring down my minus 15, or my negative 15 term.

Now I'm going to repeat that process again, only this time I'll say what will I multiply by x to give me positive 3x? That's going to be a positive 3, so I'll again write that term above. Then I am going to do step two, which is to multiply this term by both terms in my divisor. So I'm going to multiply 3 times x and 3 times negative 5 and write it underneath here. 3 times x will give me 3x, and 3 times negative 5 will give me negative 15.

Again, I'm going to subtract those two terms from the expression above it. And 3x minus 3x will again cancel out. And also, negative 15 minus negative 15 will give me zero, which is the same as those two terms also canceling out. So I'm left with zero for my remainder. And I know that my process of long division is over, because the highest power of my remainder is lower than the highest power of my divisor.

If we were to have a remainder, we would write it as the numerator of a fraction with my divisor x minus 5, the denominator of that fraction. We can also verify that x plus 3 is my solution by multiplying x plus 3 by x minus 5, and that should give me my original dividend of x squared minus 2x minus 15. So let's try that.

I'm going to multiply x plus 3 by x minus 5. Because it's two binomials, I'm going to do that by using FOIL. So my first two terms multiplied, x times x, gives me x squared. My outside terms, x and negative 5 multiplied, will give me negative 5x. My inside terms, 3 and x, will give me 3x, and my out terms, positive 3 and negative 5, will give me negative 15. I can combine my two middle terms, because they're like terms. That will give me negative 2x. And bringing my other two terms, I see that I have x squared minus 2x minus 15, which does equal my original dividend, meaning that x squared minus 2x minus 15 divided by x minus 5 is x plus 3.

For my second example, I'm going to divide the polynomial x squared plus 6x minus 7 by x minus 1. And I'm going to do that using synthetic division. The process for synthetic division starts by looking at the constant term of my divisor and using the opposite signs. So instead of negative 1, I'm going to start with a positive 1, and then I'm going to draw a box. Inside the box in a top row, I'm going to write the coefficients of my dividend. This term has a coefficient of 1. This term has a coefficient of 6. And my constant is a negative 7. So I'll write 1, 6 and negative 7.

The process for using synthetic division once you have it set up starts by bringing down the number that you have written in the first row, the first number in the first row, and writing it underneath. Then you multiply this number by the number that you had written in front initially. 1 times 1 gives me 1. And you take that value, and you write it in the second row, and then you add these two numbers together. 6 plus 1 would give me seven. And I'll write that underneath.

Then I repeat that process. I'm going to multiply 7 by 1. 7 times 1 gives me 7, and I'll write it here in the second row. Add it to the number above it-- negative 7 plus 7 will give me zero.

Now these three numbers are going to be the coefficients of my answer, which is going to be the polynomial that's the quotient of these two polynomials. So I'm going to work backwards. This last term is going to be my remainder. This is going to be my constant term, and this is going to be the coefficient of my x term. So my answer is going to be 1x, which is also just x, plus 7. And it has a remainder of zero. So x plus 7 is my answer.

For my last example, I'm going to divide the polynomial x squared plus 5x plus 4 by the polynomial x plus 3. I'm going to again use synthetic division. I'm going to start by looking at my constant term in my divisor. I notice it's a positive 3, and so I'm going to use the number with the opposite sign, negative 3. You always use the constant term with the opposite sign. So I write negative 3 and draw my box.

In the first row, I'm going to use the coefficients from the polynomial that is my dividend. So that will be 1, 5, and 4. I'll start by bringing down my first number 1 and writing it underneath, then I'll multiply 1 times negative 3, which would give me negative 3. I'll write it in the second row and then add the number above it. 5 plus negative 3 will give me a positive 2. 2 times negative 3 gives me negative 6. Negative 6 plus 4 will give me negative 2.

I know that these three numbers are going to be the coefficients of my answer. Starting from the end, I know the negative 2 will be part of my remainder. The 2 will be my constant term, and the 1 is going to be the coefficient of my x term. Writing this out, I have x plus 2, and my remainder is going to be a fraction with negative 2 in the numerator, and my divisor, x plus 3, in the denominator. So this is my final answer.

Let's go over our key points from today. When dividing polynomials, you can verify the answer, quotient, is correct by multiplying the quotient by the divisor to see that it equals the original dividend. Dividing polynomials using synthetic division involves using the coefficients of the dividend polynomial and the constant term of the divisor. Dividing polynomials using long division involves using the standard algorithm for long division.

I hope that these key points and examples helped you understand a little bit more about dividing polynomials with synthetic and long division. Keep using your notes and keep on practicing, and soon you'll be a pro. Thanks for watching.