Find the Roots (Zeros) D(x)=x^4-3x^3+x^2+3x-2

$D (x) = x^{4} - 3 x^{3} + x^{2} + 3 x - 2$

Set $x^{4} - 3 x^{3} + x^{2} + 3 x - 2$ equal to $0$ .

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

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

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

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

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

$- 3 x \cdot x^{2} + 3 x + x^{4} + x^{2} - 2 = 0$

Factor $- 3 x$ out of $3 x$ .

$- 3 x \cdot x^{2} - 3 x \cdot - 1 + x^{4} + x^{2} - 2 = 0$

Factor $- 3 x$ out of $- 3 x (x^{2}) - 3 x (- 1)$ .

$- 3 x (x^{2} - 1) + x^{4} + x^{2} - 2 = 0$

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

$- 3 x (x^{2} - 1^{2}) + x^{4} + x^{2} - 2 = 0$

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 $b = 1$ .

$- 3 x ((x + 1) (x - 1)) + x^{4} + x^{2} - 2 = 0$

Remove unnecessary parentheses.

$- 3 x (x + 1) (x - 1) + x^{4} + x^{2} - 2 = 0$

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

$- 3 x (x + 1) (x - 1) + {(x^{2})}^{2} + x^{2} - 2 = 0$

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

$- 3 x (x + 1) (x - 1) + u^{2} + u - 2 = 0$

Factor $u^{2} + u - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$- 3 x (x + 1) (x - 1) + (u - 1) (u + 2) = 0$

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

$- 3 x (x + 1) (x - 1) + (x^{2} - 1) (x^{2} + 2) = 0$

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

$- 3 x (x + 1) (x - 1) + (x^{2} - 1^{2}) (x^{2} + 2) = 0$

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 $b = 1$ .

$- 3 x (x + 1) (x - 1) + (x + 1) (x - 1) (x^{2} + 2) = 0$

Factor $(x + 1) (x - 1)$ out of $- 3 x (x + 1) (x - 1) + (x + 1) (x - 1) (x^{2} + 2)$ .

Factor $(x + 1) (x - 1)$ out of $- 3 x (x + 1) (x - 1)$ .

$(x + 1) (x - 1) (- 3 x) + (x + 1) (x - 1) (x^{2} + 2) = 0$

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) (- 3 x) + (x + 1) (x - 1) (x^{2} + 2)$ .

$(x + 1) (x - 1) (- 3 x + x^{2} + 2) = 0$

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

$(x + 1) (x - 1) (- 3 u + u^{2} + 2) = 0$

Factor $- 3 u + u^{2} + 2$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

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

Factor.

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

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

Remove unnecessary parentheses.

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

Combine exponents.

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

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

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

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

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

$x + 1 = 0$

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

$x - 2 = 0$

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 ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

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

Solve ${(x - 1)}^{2} = 0$ for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ 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 $(x + 1) {(x - 1)}^{2} (x - 2) = 0$ true.

$x = - 1, 1, 2$

Find the Roots (Zeros) D(x)=x^4-3x^3+x^2+3x-2

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