Precalculus Examples

$- 2 x^{2} + 3 x - 1$

Step 1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1$

$q = \pm 1, \pm 2$

Step 2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm \frac{1}{2}$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$- 2 {(1)}^{2} + 3 (1) - 1$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

One to any power is one.

$- 2 \cdot 1 + 3 (1) - 1$

Step 4.1.2

Multiply $- 2$ by $1$ .

$- 2 + 3 (1) - 1$

Step 4.1.3

Multiply $3$ by $1$ .

$- 2 + 3 - 1$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $- 2$ and $3$ .

$1 - 1$

Step 4.2.2

Subtract $1$ from $1$ .

$0$

Step 5

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- 2 x^{2} + 3 x - 1}{x - 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Step 6.1

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

$1$	$- 2$	$3$	$- 1$

Step 6.2

The first number in the dividend $(- 2)$ is put into the first position of the result area (below the horizontal line).

$1$	$- 2$	$3$	$- 1$

	$- 2$

Step 6.3

Multiply the newest entry in the result $(- 2)$ by the divisor $(1)$ and place the result of $(- 2)$ under the next term in the dividend $(3)$ .

$1$	$- 2$	$3$	$- 1$
		$- 2$
	$- 2$

Step 6.4

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

$1$	$- 2$	$3$	$- 1$
		$- 2$
	$- 2$	$1$

Step 6.5

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 $(- 1)$ .

$1$	$- 2$	$3$	$- 1$
		$- 2$	$1$
	$- 2$	$1$

Step 6.6

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

$1$	$- 2$	$3$	$- 1$
		$- 2$	$1$
	$- 2$	$1$	$0$

Step 6.7

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

$(- 2) x + 1$

Step 6.8

Simplify the quotient polynomial.

$- 2 x + 1$

Step 7

Solve the equation to find any remaining roots.

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Step 7.1

Subtract $1$ from both sides of the equation.

$- 2 x = - 1$

Step 7.2

Divide each term in $- 2 x = - 1$ by $- 2$ and simplify.

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Step 7.2.1

Divide each term in $- 2 x = - 1$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 1}{- 2}$

Step 7.2.2

Simplify the left side.

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Step 7.2.2.1

Cancel the common factor of $- 2$ .

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Step 7.2.2.1.1

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 1}{- 2}$

Step 7.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 2}$

Step 7.2.3

Simplify the right side.

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Step 7.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{1}{2}$

Step 8

The polynomial can be written as a set of linear factors.

$(x - 1) (x - \frac{1}{2})$

Step 9

These are the roots (zeros) of the polynomial $- 2 x^{2} + 3 x - 1$ .

$x = 1, \frac{1}{2}$

Step 10