To understand synthetic division, it helps to do the synthetic division alongside the long division problem. Note: synthetic division is used for dividing by linear functions with a leading coefficient of 1. If you have to divide by (2x-5), then first, you divide by 2, then use synthetic division to divide by x - 5/2.
Notice that on each subtraction step in long division, the first term *always* cancels out. Since you know it's going to cancel out, why bother with it in the first place?
Let's say the divisor is (x-5). Well, when you figure out the next term in the quotient, you multiply x-5 by that term. Then you subtract. I hate subtracting. When you do synthetic division, instead of having that -5 from x-5 and subtracting, you have positive 5 and you add instead.
Example:
Divide (6x^5 + 3x^3 -2x^2 +7x +11) by (x+2)
If you look at the long division problem, you know exactly what's going to happen to the x's. All you're really worrying about is the coefficients. Since adding is a bit easier than subtracting, we'll write down -2 (which is the root. Later, this turns out to be quite important.)
If you were wondering, "how many times does x+2 go into 6x^5, well, 6x^4 times. The coefficient is *always* going to be the same, and the degree is always going to be one less than the degree of that term. Writing it out in long division form just wastes time writing a lot of x's that we know will be there in the answer and what the degree will be in the answer.
So, we'd start synthetic division by writing this:
-2 | 6 0 3 -2 7 11
Notice the 0: it's a place holder for the x^4 term, since the coefficient on that term is 0
Now, we know the first term in the answer is going to have an x^4 in it. No point in writing that down; just the coefficient. And, that coefficient is going to be 6, exactly the same as the coefficient on the first term in the dividend.
So, just bring down the 6. Skip a line though.
-2 | 6 0 3 -2 7 11
......6
Now, if we were doing long division, we'd be multiplying positive 2 by the 6, then we'd be subtracting that from the 0x^4. Instead, we wrote down -2, we're still going to multiply it by 6, but we're going to add it to the 0.
-2 | 6.. 0.. 3.. -2.. 7.. 11
.........-12
......6 -12 <- so far, brought down the 6, multiplied the -2 by 6, and added it to 0
Repeat with the -12. Multiply it by the -2, and add it to the 3
-2 | 6.. 0.. 3.. -2.. 7.. 11
.........-12 24
......6 -12 27
Repeat again with the 27. In the long division, this would be 27x^3 and you'd be wondering, how many times does x go into 27x^3. 27! Then, you'd be multiplying th 27 times the +2 and subtracting. We're going to multiply it by -2 and add.
-2 | 6.. 0.. 3.. -2.. 7.. 11
.........-12 24 -48
......6 -12 27 -50
Repeat again
-2 | 6.. 0.. 3.. -2.. 7.. 11
.........-12 24 -48 100
......6 -12 27 -50 107
And again!
-2 | 6.. 0.. 3.. -2.. 7... 11
.........-12 24 -48 100 -214
......6 -12 27 -50 107 -203
So, the final answer is 6x^4 -12x^3 +27x^2 -50x +107 What the heck is that -203?? It's the remainder! So, plus -203/(x+2) is the final term.
Note: if the dividend was f(x) then f(-2) is -203. I could explain why, but that'd be another thread
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