Find the Roots (Zeros) x^4-8x^3+14x^2+8x-15=0

$x^{4} - 8 x^{3} + 14 x^{2} + 8 x - 15 = 0$

Factor the left side of the equation.

Regroup terms.

$- 8 x^{3} + 8 x + x^{4} + 14 x^{2} - 15 = 0$

Factor $- 8 x$ out of $- 8 x^{3} + 8 x$ .

Factor $- 8 x$ out of $- 8 x^{3}$ .

$- 8 x \cdot x^{2} + 8 x + x^{4} + 14 x^{2} - 15 = 0$

Factor $- 8 x$ out of $8 x$ .

$- 8 x \cdot x^{2} - 8 x \cdot - 1 + x^{4} + 14 x^{2} - 15 = 0$

Factor $- 8 x$ out of $- 8 x (x^{2}) - 8 x (- 1)$ .

$- 8 x (x^{2} - 1) + x^{4} + 14 x^{2} - 15 = 0$

Rewrite $1$ as $1^{2}$ .

$- 8 x (x^{2} - 1^{2}) + x^{4} + 14 x^{2} - 15 = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$- 8 x ((x + 1) (x - 1)) + x^{4} + 14 x^{2} - 15 = 0$

Remove unnecessary parentheses.

$- 8 x (x + 1) (x - 1) + x^{4} + 14 x^{2} - 15 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 8 x (x + 1) (x - 1) + {(x^{2})}^{2} + 14 x^{2} - 15 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 8 x (x + 1) (x - 1) + u^{2} + 14 u - 15 = 0$

Factor $u^{2} + 14 u - 15$ using the AC method.

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 $- 15$ and whose sum is $14$ .

$- 1, 15$

Write the factored form using these integers.

$- 8 x (x + 1) (x - 1) + (u - 1) (u + 15) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$- 8 x (x + 1) (x - 1) + (x^{2} - 1) (x^{2} + 15) = 0$

Rewrite $1$ as $1^{2}$ .

$- 8 x (x + 1) (x - 1) + (x^{2} - 1^{2}) (x^{2} + 15) = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$- 8 x (x + 1) (x - 1) + (x + 1) (x - 1) (x^{2} + 15) = 0$

Factor $(x + 1) (x - 1)$ out of $- 8 x (x + 1) (x - 1) + (x + 1) (x - 1) (x^{2} + 15)$ .

Factor $(x + 1) (x - 1)$ out of $- 8 x (x + 1) (x - 1)$ .

$(x + 1) (x - 1) (- 8 x) + (x + 1) (x - 1) (x^{2} + 15) = 0$

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) (- 8 x) + (x + 1) (x - 1) (x^{2} + 15)$ .

$(x + 1) (x - 1) (- 8 x + x^{2} + 15) = 0$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x + 1) (x - 1) (- 8 u + u^{2} + 15) = 0$

Factor $- 8 u + u^{2} + 15$ using the AC method.

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 $15$ and whose sum is $- 8$ .

$- 5, - 3$

Write the factored form using these integers.

$(x + 1) (x - 1) ((u - 5) (u - 3)) = 0$

Factor.

Replace all occurrences of $u$ with $x$ .

$(x + 1) (x - 1) ((x - 5) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 1) (x - 5) (x - 3) = 0$

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

$x + 1 = 0$

$x - 1 = 0$

$x - 5 = 0$

$x - 3 = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x + 1) (x - 1) (x - 5) (x - 3) = 0$ true.

$x = - 1, 1, 5, 3$

Find the Roots (Zeros) x^4-8x^3+14x^2+8x-15=0

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