Find the Roots (Zeros) P(x)=3x^4+13x^3+x^2-13x-4

$P (x) = 3 x^{4} + 13 x^{3} + x^{2} - 13 x - 4$

Step 1

Set $3 x^{4} + 13 x^{3} + x^{2} - 13 x - 4$ equal to $0$ .

$3 x^{4} + 13 x^{3} + x^{2} - 13 x - 4 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$13 x^{3} - 13 x + 3 x^{4} + x^{2} - 4 = 0$

Factor $13 x$ out of $13 x^{3} - 13 x$ .

Factor $13 x$ out of $13 x^{3}$ .

$13 x (x^{2}) - 13 x + 3 x^{4} + x^{2} - 4 = 0$

Factor $13 x$ out of $- 13 x$ .

$13 x (x^{2}) + 13 x (- 1) + 3 x^{4} + x^{2} - 4 = 0$

Factor $13 x$ out of $13 x (x^{2}) + 13 x (- 1)$ .

$13 x (x^{2} - 1) + 3 x^{4} + x^{2} - 4 = 0$

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

$13 x (x^{2} - 1^{2}) + 3 x^{4} + x^{2} - 4 = 0$

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 $b = 1$ .

$13 x ((x + 1) (x - 1)) + 3 x^{4} + x^{2} - 4 = 0$

Remove unnecessary parentheses.

$13 x (x + 1) (x - 1) + 3 x^{4} + x^{2} - 4 = 0$

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

$13 x (x + 1) (x - 1) + 3 {(x^{2})}^{2} + x^{2} - 4 = 0$

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

$13 x (x + 1) (x - 1) + 3 u^{2} + u - 4 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 4 = - 12$ and whose sum is $b = 1$ .

Multiply by $1$ .

$13 x (x + 1) (x - 1) + 3 u^{2} + 1 u - 4 = 0$

Rewrite $1$ as $- 3$ plus $4$

$13 x (x + 1) (x - 1) + 3 u^{2} + (- 3 + 4) u - 4 = 0$

Apply the distributive property.

$13 x (x + 1) (x - 1) + 3 u^{2} - 3 u + 4 u - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$13 x (x + 1) (x - 1) + 3 u^{2} - 3 u + 4 u - 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$13 x (x + 1) (x - 1) + 3 u (u - 1) + 4 (u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $u - 1$ .

$13 x (x + 1) (x - 1) + (u - 1) (3 u + 4) = 0$

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

$13 x (x + 1) (x - 1) + (x^{2} - 1) (3 x^{2} + 4) = 0$

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

$13 x (x + 1) (x - 1) + (x^{2} - 1^{2}) (3 x^{2} + 4) = 0$

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 $b = 1$ .

$13 x (x + 1) (x - 1) + (x + 1) (x - 1) (3 x^{2} + 4) = 0$

Factor $(x + 1) (x - 1)$ out of $13 x (x + 1) (x - 1) + (x + 1) (x - 1) (3 x^{2} + 4)$ .

Factor $(x + 1) (x - 1)$ out of $13 x (x + 1) (x - 1)$ .

$(x + 1) (x - 1) (13 x) + (x + 1) (x - 1) (3 x^{2} + 4) = 0$

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) (13 x) + (x + 1) (x - 1) (3 x^{2} + 4)$ .

$(x + 1) (x - 1) (13 x + 3 x^{2} + 4) = 0$

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

$(x + 1) (x - 1) (13 u + 3 u^{2} + 4) = 0$

Factor by grouping.

Reorder terms.

$(x + 1) (x - 1) (3 u^{2} + 13 u + 4) = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 4 = 12$ and whose sum is $b = 13$ .

Factor $13$ out of $13 u$ .

$(x + 1) (x - 1) (3 u^{2} + 13 (u) + 4) = 0$

Rewrite $13$ as $1$ plus $12$

$(x + 1) (x - 1) (3 u^{2} + (1 + 12) u + 4) = 0$

Apply the distributive property.

$(x + 1) (x - 1) (3 u^{2} + 1 u + 12 u + 4) = 0$

Multiply $u$ by $1$ .

$(x + 1) (x - 1) (3 u^{2} + u + 12 u + 4) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) (x - 1) ((3 u^{2} + u) + 12 u + 4) = 0$

Factor out the greatest common factor (GCF) from each group.

$(x + 1) (x - 1) (u (3 u + 1) + 4 (3 u + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 u + 1$ .

$(x + 1) (x - 1) ((3 u + 1) (u + 4)) = 0$

Factor.

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

$(x + 1) (x - 1) ((3 x + 1) (x + 4)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 1) (3 x + 1) (x + 4) = 0$

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

$x + 1 = 0$

$x - 1 = 0$

$3 x + 1 = 0$

$x + 4 = 0$

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 $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = - 1, 1, - \frac{1}{3}, - 4$

Step 3

Find the Roots (Zeros) P(x)=3x^4+13x^3+x^2-13x-4

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