Use the Factor Theorem to Determine if x=1 is a Factor f(x)=8x-3 , x=1

$f (x) = 8 x - 3$ , $x = 1$

Set up the long division problem to evaluate the function at $1$ .

$\frac{8 x - 3}{x - (1)}$

Divide using synthetic division.

Place the numbers representing the divisor and the dividend into a division-like configuration.

$1$	$8$	$- 3$
	\xa0	\xa0

The first number in the dividend $(8)$ is put into the first position of the result area (below the horizontal line).

Multiply the newest entry in the result $(8)$ by the divisor $(1)$ and place the result of $(8)$ under the next term in the dividend $(- 3)$ .

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$8 + \frac{5}{x - 1}$

The remainder of the synthetic division is the result based on the remainder theorem.

$5$

Since the remainder is not equal to zero, $x = 1$ is not a factor.

$x = 1$ is not a factor