6.4 Division of Polynomials. (Long Division and Synthetic Division)

Transcription

1 6.4 Division of Polynomials (Long Division and Synthetic Division)

2 When we combine fractions that have a common denominator, we just add or subtract the numerators and then keep the common denominator on our combined fraction. For dividing, we do just the opposite. We separate the fraction into individual parts by writing each term in the numerator by itself and then put the denominator under each one as shown below. Divide: 6.

3 Let's review regular long division with just numbers: Divisor Quotient Remainder divided by divisor Side Work Remainder

4 (We will do this problem first with long division of polynomials and then with synthetic division.) We use the same technique in polynomial long division as we do in regular number long division, except instead of thinking "how many times does the divisor go into the number," we think "what do we multiply just the first term of the divisor by to get the first term of the polynomial we are dividing it into". Note: One thing you always have to do first is make sure all polynomials are in standard form (highest power to lowest power). Divide: 10. Long Division Method: Side Work: Multiply it by the entire divisor Multiply it by the entire divisor Notice that on each step when we subtract, we put parenthesis around the entire expression to subtract. Since adding is easier than subtracting, we then changed the "subtract" to "add" the opposite signs. Synthetic Division Method: For synthetic division, we put just the coefficients inside the upside down division bar, and we put the zero of the divisor on the outside. We use the zero because it has the opposite sign of the factor (the divisor), so instead of subtracting (like we did on each step of long division) we can do the opposite and just add now. First drop the first number inside the bar down. Then multiply diagonally ( 5 times 3) and put it under the next number inside the bar (under the 19). Then add straight down ( ) and put that answer (4) right below under the bar. Then do the process again ( 5 times 4 = 20, so put the 20 under the 20 inside the bar and then add straight down to get zero under the bar.) Now we have under the bar. Since we started with 3x 2 as the first term of the polynomial we divided into and we divided out one x, we now have x1 as the variable for the answer and we go down in powers as we go to the right as follows. Therefore: is actually 3x + 4 which is the final answer (the quotient).

5 Divide: 14. Long Division Method: Side Work: Synthetic Division Method: This time the zero of the divisor (4x + 8) is 2 (because solving 4x + 8 = 0 gives us x = 2), so we put a 2 on the outside of the bar. Since our first term was 8x 3, our answer (quotient) with start with an x 2 term. Notice that is quotient is not the same as the quotient we got when we did the long division. The reason is because the divisor (4x + 8) has a coefficient other than 1 on the x term. Basically, when we just divided using the zero of 2, we ignored that we really needed to divide by 4 also, [4(x + 2)]. So our answer using synthetic division is not correct. We need to divide it by 4 and then we will get the same answer we did using long division. And reduce each term:

6 Divide: 16. First reorder the terms into standard form (highest power to lowest power). Solve 2x 3 = 0 to get the zero of 3/2. Side Work: Divide the final answer by 2 because our divisor was actually (2x 3) [or 2(x 3/2)].

7 This problem is in standard form, but it is missing a term in the polynomial we are dividing into, so we must insert 0x 2 term in it just as a place holder. Then when we subtract any x 2 terms, we have a place to put them. Divide: Side Work: 18. (I added +0 (in red) just so I'd have a place holder for the missing terms.) *Note that this time we have a remainder that is not zero, so we add to the end of the quotient: remainder divided by divisor.

8 This example is not in your book. When we are asked to find P(4), we know that we just plug 4 in for every x in P(x), but as shown in the problem below, this can often involve very large numbers which make it more difficult to find. Luckily, there is another way to find P(4) using synthetic division. Using the method of just plugging in 4 for every x: Find P(4): = Now find P(4) using synthetic division: Notice that we put 4 (not 4) outside the division bar. Remember, we are not actually doing polynomial division here we are just finding P(4). the answer to this problem is just 19, not the entire quotient polynomial. This method is so much easier! The remainder is P(4). P(4) = 19