Originally Posted by

**algebraisabeast** If I am simplifing an expression in which I cannot factor I can use polynomial division, I can also use synethic division to get the same answer. I know that I can only use synethic division when I am dividing by a linear equation.

When I change the leading coefficient on the linear equation I found that I have to divide my numbers when I do synthetic division by that number, why does this have to happen

example using poly division

(42x^2-33)/(7x+7) = 6x-6+9/7x+7

using synthetic I get 42x-42 with a remainder of 9

so I can say that r= 9/x+1 Is the reason that I divide by 7 because I take that out a 7 and use -1 so I have to put the 7 back in so I have to divide everything by 7 to get the same answer when I use poly division?

If I am using synthetic division to solve my answer will I always divide my two numbers by the leading cooefficient on my linear term.

Hi algebraisabeast,

Your polynomial long division answer is correct:

$\displaystyle (42x^2-33) \div (7x+7) = 6x-6+\frac{9}{7x+7}$

Now, when you attempt to use synthetic division and the coefficient of x is not 1, then you must divide everything by that coefficient, the divisor and the dividend.

Dividing everything by 7 you would have $\displaystyle (6x^2-\frac{33}{7}) \div (x+1)$

Now set up your synthetic division process. Your divisor is now -1.

Code:

-1 | 6 0 -33/7
-6 6
-------------
6 -6 9/7

Above translates to $\displaystyle 6x-6+\frac{\frac{9}{7}}{x+1}$ which further simplifies to

$\displaystyle 6x-6+\frac{9}{7x+7}$

Originally Posted by

**algebraisabeast** like if I want to divide 2x^2-17x-38 by 2x+3 if I use synthetic I will just divide by 2 when I am done

50k^3+10k^2-35k-7 divided by 5k-4 when I do synthetic divide all terms by 5?

This is correct. Now try one of those and see if you can do it.