Precalculus Examples

$x - 1$ , $x^{2} + 2 x + 5$

Step 1

Divide $\frac{x^{2} + 2 x + 5}{x - 1}$ using synthetic division and check if the remainder is equal to $0$ . If the remainder is equal to $0$ , it means that $x - 1$ is a factor for $x^{2} + 2 x + 5$ . If the remainder is not equal to $0$ , it means that $x - 1$ is not a factor for $x^{2} + 2 x + 5$ .

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Step 1.1

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

$1$	$1$	$2$	$5$

Step 1.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$1$	$1$	$2$	$5$

	$1$

Step 1.3

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(2)$ .

$1$	$1$	$2$	$5$
		$1$
	$1$

Step 1.4

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

$1$	$1$	$2$	$5$
		$1$
	$1$	$3$

Step 1.5

Multiply the newest entry in the result $(3)$ by the divisor $(1)$ and place the result of $(3)$ under the next term in the dividend $(5)$ .

$1$	$1$	$2$	$5$
		$1$	$3$
	$1$	$3$

Step 1.6

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

$1$	$1$	$2$	$5$
		$1$	$3$
	$1$	$3$	$8$

Step 1.7

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

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

Step 1.8

Simplify the quotient polynomial.

$x + 3 + \frac{8}{x - 1}$

Step 2

The remainder from dividing $\frac{x^{2} + 2 x + 5}{x - 1}$ is $8$ , which is not equal to $0$ . The remainder is not equal to $0$ means that $x - 1$ is not a factor for $x^{2} + 2 x + 5$ .

$x - 1$ is not a factor for $x^{2} + 2 x + 5$