Find the Roots (Zeros) x^5-3x^4+x^3+x^2+4=0

$x^{5} - 3 x^{4} + x^{3} + x^{2} + 4 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$x^{5} + x^{2} - 3 x^{4} + x^{3} + 4 = 0$

Factor $x^{2}$ out of $x^{5} + x^{2}$ .

Factor $x^{2}$ out of $x^{5}$ .

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

Multiply by $1$ .

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

Factor $x^{2}$ out of $x^{2} x^{3} + x^{2} \cdot 1$ .

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

Rewrite $1$ as $1^{3}$ .

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

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Factor.

Simplify.

Multiply $- 1$ by $1$ .

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

One to any power is one.

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

Remove unnecessary parentheses.

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

Factor $- 3 x^{4} + x^{3} + 4$ using the rational roots test.

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.

$p = \pm 1, \pm 4, \pm 2 q = \pm 1, \pm 3$

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

$\pm 1, \pm 0.\overline{3}, \pm 4, \pm 1.\overline{3}, \pm 2, \pm 0.\overline{6}$

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

Substitute $- 1$ into the polynomial.

$- 3 {(- 1)}^{4} + {(- 1)}^{3} + 4$

Raise $- 1$ to the power of $4$ .

$- 3 \cdot 1 + {(- 1)}^{3} + 4$

Multiply $- 3$ by $1$ .

$- 3 + {(- 1)}^{3} + 4$

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

$- 3 - 1 + 4$

Subtract $1$ from $- 3$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

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.

$\frac{- 3 x^{4} + x^{3} + 4}{x + 1}$

Divide $- 3 x^{4} + x^{3} + 4$ by $x + 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

Divide the highest order term in the dividend $- 3 x^{4}$ by the highest order term in divisor $x$ .

$3 x^{3}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

Multiply the new quotient term by the divisor.

$3 x^{3}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $- 3 x^{4} - 3 x^{3}$

$3 x^{3}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$3 x^{3}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

Pull the next terms from the original dividend down into the current dividend.

$3 x^{3}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

Divide the highest order term in the dividend $4 x^{3}$ by the highest order term in divisor $x$ .

$3 x^{3}$

$4 x^{2}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

Multiply the new quotient term by the divisor.

$3 x^{3}$

$4 x^{2}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{3} + 4 x^{2}$

$3 x^{3}$

$4 x^{2}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$3 x^{3}$

$4 x^{2}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

$3 x^{3}$

$4 x^{2}$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

$3 x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

Multiply the new quotient term by the divisor.

$3 x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} - 4 x$

$3 x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$3 x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

Pull the next terms from the original dividend down into the current dividend.

$3 x^{3}$

$4 x^{2}$

$4 x$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

$4$

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

$3 x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

$4$

Multiply the new quotient term by the divisor.

$3 x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

$4$

$4 x$

$4$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x + 4$

$3 x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

$4$

$4 x$

$4$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$3 x^{3}$

$4 x^{2}$

$4 x$

$4$

$x$

$1$

$3 x^{4}$

$x^{3}$

$0 x^{2}$

$0 x$

$4$

$3 x^{4}$

$3 x^{3}$

$4 x^{3}$

$0 x^{2}$

$4 x^{3}$

$4 x^{2}$

$0 x$

$4 x^{2}$

$4 x$

$4$

$4 x$

$4$

$0$

Since the remander is $0$ , the final answer is the quotient.

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

Write $- 3 x^{4} + x^{3} + 4$ as a set of factors.

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

Factor $x + 1$ out of $x^{2} (x + 1) (x^{2} - x + 1) + (x + 1) (- 3 x^{3} + 4 x^{2} - 4 x + 4)$ .

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

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

Factor $x + 1$ out of $(x + 1) (x^{2} (x^{2} - x + 1)) + (x + 1) (- 3 x^{3} + 4 x^{2} - 4 x + 4)$ .

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

Apply the distributive property.

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

Simplify.

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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

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

Add $2$ and $2$ .

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

Rewrite using the commutative property of multiplication.

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

Multiply $x^{2}$ by $1$ .

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

Multiply $x^{2}$ by $x$ by adding the exponents.

Move $x$ .

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

Multiply $x$ by $x^{2}$ .

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

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

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

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

Add $1$ and $2$ .

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

Subtract $3 x^{3}$ from $- x^{3}$ .

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

Add $x^{2}$ and $4 x^{2}$ .

$(x + 1) (x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4) = 0$

Step 2

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

$x + 1 = 0$

$x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4 = 0$

Step 3

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.

$x = - 1$

Step 4

Set $x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4$ equal to $0$ and solve for $x$ .

Set $x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4$ equal to $0$ .

$x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4 = 0$

Solve $x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4 = 0$ for $x$ .

Factor the left side of the equation.

Regroup terms.

$- 4 x^{3} - 4 x + x^{4} + 5 x^{2} + 4 = 0$

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

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

$- 4 x \cdot x^{2} - 4 x + x^{4} + 5 x^{2} + 4 = 0$

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

$- 4 x \cdot x^{2} - 4 x \cdot 1 + x^{4} + 5 x^{2} + 4 = 0$

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

$- 4 x (x^{2} + 1) + x^{4} + 5 x^{2} + 4 = 0$

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

$- 4 x (x^{2} + 1) + {(x^{2})}^{2} + 5 x^{2} + 4 = 0$

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

$- 4 x (x^{2} + 1) + u^{2} + 5 u + 4 = 0$

Factor $u^{2} + 5 u + 4$ 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 $5$ .

$1, 4$

Write the factored form using these integers.

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

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

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

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

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

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

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

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

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

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

Factor using the perfect square rule.

Rearrange terms.

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

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

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

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.

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

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

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

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

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

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

$x^{2} + 1 = 0$

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

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

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

$x = i, - i, 2$

Step 5

The final solution is all the values that make $(x + 1) (x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4) = 0$ true.

$x = - 1, i, - i, 2$

Step 6

Find the Roots (Zeros) x^5-3x^4+x^3+x^2+4=0

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