Synthetic Division | MathHelp.com - By MathHelp.com
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| 00:0-1 | in this example notice that we have a polynomial divided | |
| 00:03 | by a binomial and are by no meal is in | |
| 00:06 | the form of an X term minus a constant term | |
| 00:11 | or x minus C . In this situation , instead | |
| 00:15 | of having to use long division like we did in | |
| 00:18 | the previous lesson we can divide the polynomial is using | |
| 00:23 | synthetic division , which is a much more efficient method | |
| 00:28 | . Here's how it works . We start by finding | |
| 00:31 | the value of C since negative C equals negative three | |
| 00:37 | . We know that C equals three . Next we | |
| 00:42 | put the value of C inside a box . So | |
| 00:45 | we put the three inside a box . It's very | |
| 00:49 | important to understand that the number that goes inside the | |
| 00:53 | box always uses the opposite sign of the constant term | |
| 00:58 | in the binomial . In other words , since the | |
| 01:02 | constant term in the binomial is negative three , The | |
| 01:06 | number that goes inside the box is positive three . | |
| 01:11 | Next we write the coefficients of the dividend Which are | |
| 01:16 | too negative 7 , 4 and five . Be very | |
| 01:21 | careful with your signs . Now we're ready to start | |
| 01:25 | our synthetic division 1st . We bring down the two | |
| 01:31 | . Next we multiply the three in the box times | |
| 01:34 | two to get six And we put the six under | |
| 01:39 | the -7 . Next we add negative seven plus six | |
| 01:44 | to get negative one next we multiply the three in | |
| 01:48 | the box times negative one to get negative three . | |
| 01:52 | And we put the -3 under the four . Next | |
| 01:57 | we add four plus negative three to get one . | |
| 02:01 | Next we multiply the three in the box times one | |
| 02:05 | to get three And we put the three under the | |
| 02:08 | five . Finally we add five plus three to get | |
| 02:12 | eight . Now notice that we have a two negative | |
| 02:17 | 11 and eight in the bottom row of our synthetic | |
| 02:22 | division . These values will give us our answer . | |
| 02:27 | The first three numbers represent the coefficients of the quotation | |
| 02:32 | and the last number is the remainder . And it's | |
| 02:35 | important to understand that our answer will be one degree | |
| 02:39 | less than the dividend . In other words , since | |
| 02:43 | our dividend starts with X cubed and we're dividing by | |
| 02:48 | X . Our answer will start with X squared , | |
| 02:53 | so our answer is two X squared minus one , | |
| 02:57 | X plus one Plus eight Over X -3 . Notice | |
| 03:03 | that we always use descending order of powers in our | |
| 03:07 | quotation in this case X squared X . And the | |
| 03:12 | constant . Finally remember that we add the remainder over | |
| 03:16 | the divisor just like we did in the previous lesson | |
| 03:20 | on long division and we have our answer . It's | |
| 03:25 | important to understand that will get the same answer whether | |
| 03:29 | we use synthetic division or long division . However , | |
| 03:33 | synthetic division is much faster . |
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