# How do you use synthetic division to see if -1 is a zero of h(x)=x^4+9x^3+18x-8?

Jul 23, 2015

$- 1$ is not a zero of $h \left(x\right) = {x}^{4} + 9 {x}^{3} + 18 x - 8$

#### Explanation:

Step 1. Write only the coefficients of $x$ inside an upside-down division symbol.

Step 2. Put the test zero, $x = - 1$, at the left.

Step 3. Drop the first coefficient below the division symbol.

Step 4. Multiply the drop-down by the test zero, and put the result in the next column.

Step 5. Add down the column.

Step 6. Repeat Steps 4 and 5 until you can go no farther.

$- 1$ is not a zero of $h \left(x\right)$ because the division gives a remainder of $- 34$.

Check:

$\left(x + 1\right) \left({x}^{3} + 8 {x}^{2} - 8 x + 26 - \frac{34}{x + 1}\right)$

$= \left(x + 1\right) \left({x}^{3} + 8 {x}^{2} - 8 x + 26\right) - 34$

$= {x}^{4} + 8 {x}^{3} - \cancel{8 {x}^{2}} + 26 x + {x}^{3} + \cancel{8 {x}^{2}} - 8 x + 26 - 34$

$= {x}^{4} + 9 {x}^{3} + 18 x - 8$