# How do you use synthetic division to divide (4x^3+16x^2-23x-15)/(x+1/2)?

Aug 28, 2017

See below.

#### Explanation:

First, determine your divisor, which will be the zero of the factor you're given. (It may or may not be a zero, but I'll discuss that later.) In this case, you are dividing by $x + \frac{1}{2}$, so the divisor is $- \frac{1}{2}$. I put this in the box on the left. Next, bring down the coefficients of the dividend. Since the dividend is $4 {x}^{3} + 16 {x}^{2} - 23 x - 15$, the coefficients are $4 , 16 , - 23 ,$ and $- 15$.

Now, bring down the first coefficient, $4$, and place it below the line. Multiply $4$ by the divisor, $- \frac{1}{2}$, to get $- 2$ and put it below the second coefficient, $16$. Now add $16$ to $- 2$ to get $14$. Repeat these steps until you reach the last coefficient.

$4 \cdot - \frac{1}{2} = - 2 \to 16 + - 2 = 14$

$14 \cdot - \frac{1}{2} = - 7 \to - 23 + - 7 = - 30$

$- 30 \cdot - \frac{1}{2} = 15 \to - 15 + 15 = 0$

In the end, we have four numbers: $4 , 14 , - 30 , 0$. These are the coefficients for the quotient. Since the problem asked us to divide $4 {x}^{3} + 16 {x}^{2} - 23 x - 15$ (cubic) by $x + \frac{1}{2}$ (linear), the quotient will be quadratic. Thus, the degree of the first term will be $2$, the degree of the second term will be $1$, the degree of the third term will be $0$ (the third term is a constant), and the last term will be the remainder.

$\textcolor{w h i t e}{I}$
Since the last term in this case is $0$, there is no remainder. This tells us that $x + \frac{1}{2}$ is a factor of $4 {x}^{3} + 16 {x}^{2} - 23 x - 15$ because it divides evenly. If there was a remainder, the $x = - \frac{1}{2}$ would not be a zero, and $x + \frac{1}{2}$ would not be a factor.

In conclusion, our quotient is $\textcolor{b l u e}{4 {x}^{2} + 14 x - 30}$.