Identify the Zeros and Their Multiplicities (x^3-5x^2+2x+8)/(x+1)

$\frac{(x^{3} - 5 x^{2} + 2 x + 8)}{(x + 1)}$

To find the roots/zeros, set $\frac{x^{3} - 5 x^{2} + 2 x + 8}{x + 1}$ equal to $0$ and solve.

$\frac{x^{3} - 5 x^{2} + 2 x + 8}{x + 1} = 0$

Find the LCD of the terms in the equation.

Tap for more steps…

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x + 1, 1$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

The LCM of $x + 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x + 1$

Multiply each term by $x + 1$ and simplify.

Tap for more steps…

Multiply each term in $\frac{x^{3} - 5 x^{2} + 2 x + 8}{x + 1} = 0$ by $x + 1$ in order to remove all the denominators from the equation.

$\frac{x^{3} - 5 x^{2} + 2 x + 8}{x + 1} \cdot (x + 1) = 0 \cdot (x + 1)$

Cancel the common factor of $x + 1$ .

Tap for more steps…

Cancel the common factor.

$\frac{x^{3} - 5 x^{2} + 2 x + 8}{x + 1} \cdot (x + 1) = 0 \cdot (x + 1)$

Rewrite the expression.

$x^{3} - 5 x^{2} + 2 x + 8 = 0 \cdot (x + 1)$

Multiply $0$ by $x + 1$ .

$x^{3} - 5 x^{2} + 2 x + 8 = 0$

Solve the equation.

Tap for more steps…

Factor the left side of the equation.

Tap for more steps…

Factor $x^{3} - 5 x^{2} + 2 x + 8$ using the rational roots test.

Tap for more steps…

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.

$p = \pm 1, \pm 8, \pm 2, \pm 4 q = \pm 1$

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

$\pm 1, \pm 8, \pm 2, \pm 4$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Tap for more steps…

Substitute $- 1$ into the polynomial.

${(- 1)}^{3} - 5 {(- 1)}^{2} + 2 \cdot - 1 + 8$

Raise $- 1$ to the power of $3$ .

$- 1 - 5 {(- 1)}^{2} + 2 \cdot - 1 + 8$

Raise $- 1$ to the power of $2$ .

$- 1 - 5 \cdot 1 + 2 \cdot - 1 + 8$

Multiply $- 5$ by $1$ .

$- 1 - 5 + 2 \cdot - 1 + 8$

Subtract $5$ from $- 1$ .

$- 6 + 2 \cdot - 1 + 8$

Multiply $2$ by $- 1$ .

$- 6 - 2 + 8$

Subtract $2$ from $- 6$ .

$- 8 + 8$

Add $- 8$ and $8$ .

$0$

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.

$\frac{x^{3} - 5 x^{2} + 2 x + 8}{x + 1}$

Divide $x^{3} - 5 x^{2} + 2 x + 8$ by $x + 1$ .

$x^{2} - 6 x + 8$

Write $x^{3} - 5 x^{2} + 2 x + 8$ as a set of factors.

$(x + 1) (x^{2} - 6 x + 8) = 0$

Factor $x^{2} - 6 x + 8$ using the AC method.

Tap for more steps…

Factor $x^{2} - 6 x + 8$ using the AC method.

Tap for more steps…

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$(x + 1) ((x - 4) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 4) (x - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 1 = 0$

$x - 4 = 0$

$x - 2 = 0$

Set the first factor equal to $0$ and solve.

Tap for more steps…

Set the first factor equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set the next factor equal to $0$ and solve.

Tap for more steps…

Set the next factor equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set the next factor equal to $0$ and solve.

Tap for more steps…

Set the next factor equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x + 1) (x - 4) (x - 2) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - 1$ (Multiplicity of $1$ )

$x = 4$ (Multiplicity of $1$ )

$x = 2$ (Multiplicity of $1$ )

$x = - 1$ (Multiplicity of $1$ )

$x = 4$ (Multiplicity of $1$ )

$x = 2$ (Multiplicity of $1$ )

Identify the Zeros and Their Multiplicities (x^3-5x^2+2x+8)/(x+1)