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

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Answer 1 · 2015-07-23T23:26:20

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

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

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

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

Check:

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

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

$= x^4+8x^3-cancel(8x^2)+26x+x^3+cancel(8x^2)-8x+26 -34$

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

How do you use synthetic division to see if -1 is a zero of h(x)=x^4+9x^3+18x-8h(x)=x4+9x3+18x−8h(x)=x^4+9x^3+18x-8?