“I wish Aunt Sally was dead!” – Mr. Bieniek 4/20/2009

Today in Honors Precalculus, we were exposed to Synthetic Substitution. Synthetic Substitution is a shortcut for Polynomial Division. However, it’s not used divide out Factors, but to find Zeros. The main concept behind Synthetic Substitution is the multiply by x, then add the next coefficient property.

So, for the problem `f(x)=x^3-4x^2-3x+18` you would start by putting the coefficients starting from the greatest multiple of x. In this situation the first coefficient will be 1, which is from `x^3` or `1x^3`. The coefficient will be -4, which is from `4x^2`. Just keep on doing this until you end up with the last number of a plus 18. You numbers should be:

1     -4     -3     18

Next, you need to either find a zero of the equation or you’ll be given one. In this situation you are given that one of the zeros is x=-2 which is the next part to put into the synthetic substitution formula.  The -2 goes in front of all the previous numbers, however it’s not part of that same group. If you need a visual reminder feel free to circle or box out the zero. It should look like this.

-2 1     -4     -3     18

Now you are ready to use synthetic substitution. Draw a line underneath those numbers, but give it some room so you can write more numbers underneath. Starting with the 1 you just carry it down. What you are doing is addition so 1+0=1. The 0 is from it being the first addition you do in the problem. Next, you take your zero of the equation, and multiply that and the 1 you just put down. This gives you -2 which you put right underneath your next coefficient.

-2 1     -4     -3     18

0      -2

____________________________

1(-2)

What you’re doing is the mulitply by x and then add the next coefficient property. Just multiply your zero by the coefficent you’re adding until you run out of coefficients. The final product should look like this:

-2 1     -4     -3     18

0     -2     12     -18

__________________

1     -6       9      0

If you get a 0 at the end, that means that the number you multiplied by was indeed a zero of the equation. The numbers you get at the end of this are your coefficients for your next equation to derive the zeros from.

`f(x)=(x+2)(x^2-6x+9)`

Finally, all you have left to do is solve for the zeros. All you have to do is use your quadratic formula for this.

`f(x)=(-6+(-6^2-4(*9)))/2` Which equals -3. Then you solve for the other one by solving for `f(x)=(-6-(-6^2-4(*9)))/2` which is 3.

Your zeros are -2,-3, and 3.

By the way this is a link to the Fundamental Theorem of Algebra:

http://www.cut-the-knot.org/do_you_know/fundamental2.shtml

We discussed this is class, and this website has done a great job of explaining it.