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

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

Step 1

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

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

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

Factor $- 4 x$ out of $- 4 x^{3} + 4 x$ .

Factor $- 4 x$ out of $- 4 x^{3}$ .

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

Factor $- 4 x$ out of $4 x$ .

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

Factor $- 4 x$ out of $- 4 x (x^{2}) - 4 x (- 1)$ .

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

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

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

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 $b = 1$ .

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

Remove unnecessary parentheses.

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

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

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

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

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

Factor $u^{2} + 3 u - 4$ 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

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

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

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

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

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

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 $b = 1$ .

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

Factor $(x + 1) (x - 1)$ out of $- 4 x (x + 1) (x - 1) + (x + 1) (x - 1) (x^{2} + 4)$ .

Factor $(x + 1) (x - 1)$ out of $- 4 x (x + 1) (x - 1)$ .

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

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) (- 4 x) + (x + 1) (x - 1) (x^{2} + 4)$ .

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

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

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

Factor using the perfect square rule.

Rearrange terms.

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

Rewrite $4$ as $2^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 u = 2 \cdot u \cdot 2$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 2$ .

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

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

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

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

$x + 1 = 0$

$x - 1 = 0$

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

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

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

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

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

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

$x = - 1, 1, 2$

Step 3

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

Related Questions

Support LCMGCF.com by disabling your adblocker.