I usually know how to do Synthetic Division, but this one has my stumped. I'm not sure how to do it with the given divisor.
Here's the equation
(2x^3-17z^2+20x+6)/(2x-5)
Synthetic division can only be done when the coefficient of x is 1. So:
$\displaystyle \frac{2x^3 - 17x^2 + 20x + 6}{2x - 5}$
$\displaystyle = \frac{\frac{2x^3 - 17x^2 + 20x + 6}{2}}{\frac{2x - 5}{2}}$
$\displaystyle = \frac{x^3 - \frac{17}{2}x^2 + 10x + 3}{x - \frac{5}{2}}$
So divide $\displaystyle x^3 - \frac{17}{2}x^2 + 10x + 3$ by $\displaystyle
x - \frac{5}{2}$.
-Dan
Hello, Tony!
You should have been shown how to handle these . . .
The divisor must have a leading coefficient of 1.$\displaystyle (2x^3-17z^2+20x+6) \div (2x-5)$
. . If the leading coefficient isn't 1, divide through by it.
So we have: .$\displaystyle 2x - 5 \quad\Rightarrow\quad x - \frac{5}{2}$
We will use $\displaystyle \frac{5}{2}$ for our divisor.
Important!
Since we are dividing by one-half of the given divisor,
. . the quotient will be twice as large as it should be.
We will make the adjustment at the end.
The synthetic division looks like this:
. . . $\displaystyle \begin{array}{cccccc}
\frac{5}{2} & | & 2 & \text{-}17 & 20 & 6 \\
& | & & 5 & \text{-}30 & \text{-}25 \\
& & -- & -- & -- & -- \\
& & 2 & \text{-}12 & \text{-}10 & \text{-}19 \end{array}$
The bottom says: .$\displaystyle 2x^2 - 12x - 10,\quad \text{rem -}19$
But the quotient is twice as large as it should be ... remember?
. . The remainder is unaffected.
Therefore, the answer is: . $\displaystyle x^2 - 6x - 5 -\frac{19}{2x-5} $