Thread: Synthetic Division

Alright, i have to discuss how synthetic division can be used with any linear factor of the form ax+b. for example, any polynomial divided synthetically by ax+b. I didn't know that it was possible i thought you could only do x+b or x-b with no leading coefficient. How would you do it with any leading coefficiant. A probelm like: 4x^3-8x^2-11x+9 / 2x+3

I do not know exactly what you are asking, but if you have a polynomial of degree then by synthetic division (assuming the remainder polynomial is zero) will have degree . What you can do with this is the following useful trick.
Given a polynomial:

divide it synthetically by where is a zero of the polynomial. Your polynomial must reduce to:
.
Now to find the zeros of all need now is to find zeros of . This is especially useful when you are solving a cubic. Because the steps for solving a cubic a rather long and complicated if you can spot a trivial solution you can syntheticaly divide which will leave you with a quadradic equation. Now apply the quadratic formula and you have your 3 roots of the cubic.

Originally Posted by aussiekid90

Alright, i have to discuss how synthetic division can be used with any linear factor of the form ax+b. for example, any polynomial divided synthetically by ax+b. I didn't know that it was possible i thought you could only do x+b or x-b with no leading coefficient. How would you do it with any leading coefficiant. A probelm like: 4x^3-8x^2-11x+9 / 2x+3

I have not studied before how to do synthetic division using (ax +b). It was by (x +a) only. So I'd reduce the (ax +b) into (x +c), then do it as usual.
In doing that, I'd divide all by the coefficient of the x in the (ax +b).

(4x^3 -8x^2 -11x +9) / (2x +3)
I'd divide everything by 2,
(2x^3 -4x^2 -5.5x +4.5) / (x +1.5)
and do the usual synthetic division.

..-1.5..|....2....-4....-5.5....4.5
.....................-3....10.5...-7.5
_____________________________
...............2....-7.....5.......-3 <----there is a remainder of (-3). That means (2x +3) is not a factor of 4x^3 -8x^2 -11x +9.

Originally Posted by ticbol

I have not studied before how to do synthetic division using (ax +b). It was by (x +a) only. So I'd reduce the (ax +b) into (x +c), then do it as usual.
In doing that, I'd divide all by the coefficient of the x in the (ax +b).

(4x^3 -8x^2 -11x +9) / (2x +3)
I'd divide everything by 2,
(2x^3 -4x^2 -5.5x +4.5) / (x +1.5)
and do the usual synthetic division.

..-1.5..|....2....-4....-5.5....4.5
.....................-3....10.5...-7.5
_____________________________
...............2....-7.....5.......-3 <----there is a remainder of (-3). That means (2x +3) is not a factor of 4x^3 -8x^2 -11x +9.

If you really want to you can express

as

Thus your assertation that does not divide was correct. You reasoning by reducing to then syntheticaly dividing lead to the correct solution that it does not divide it. I just divided it out without using your step. Meaning first divided by then multiplied result and subtracted... the entire synthetic algorithm but just with

Originally Posted by ThePerfectHacker

If you really want to you can express

as

Thus your assertation that does not divide was correct. You reasoning by reducing to then syntheticaly dividing lead to the correct solution that it does not divide it. I just divided it out without using your step. Meaning first divided by then multiplied result and subtracted... the entire synthetic algorithm but just with

What synthetic algorithm? What you did was long division, not synthetic division.

Can you show how to do synthetic division using the (2x+3)? Let us see it.

There is a difference between synthetic divison and long division?

Originally Posted by ThePerfectHacker

There is a difference between synthetic divison and long division?

Yes. Synthetic division writes out the coefficients of the polynomial terms and uses a shortcut method to arrive at the quotient.