Find the Roots (Zeros) f(x)=7x^4-6x^3+27x^2-24x-4

$f (x) = 7 x^{4} - 6 x^{3} + 27 x^{2} - 24 x - 4$

Step 1

Set $7 x^{4} - 6 x^{3} + 27 x^{2} - 24 x - 4$ equal to $0$ .

$7 x^{4} - 6 x^{3} + 27 x^{2} - 24 x - 4 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$- 6 x^{3} - 24 x + 7 x^{4} + 27 x^{2} - 4 = 0$

Factor $- 6 x$ out of $- 6 x^{3} - 24 x$ .

Factor $- 6 x$ out of $- 6 x^{3}$ .

$- 6 x \cdot x^{2} - 24 x + 7 x^{4} + 27 x^{2} - 4 = 0$

Factor $- 6 x$ out of $- 24 x$ .

$- 6 x \cdot x^{2} - 6 x \cdot 4 + 7 x^{4} + 27 x^{2} - 4 = 0$

Factor $- 6 x$ out of $- 6 x (x^{2}) - 6 x (4)$ .

$- 6 x (x^{2} + 4) + 7 x^{4} + 27 x^{2} - 4 = 0$

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

$- 6 x (x^{2} + 4) + 7 {(x^{2})}^{2} + 27 x^{2} - 4 = 0$

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

$- 6 x (x^{2} + 4) + 7 u^{2} + 27 u - 4 = 0$

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 $a \cdot c = 7 \cdot - 4 = - 28$ and whose sum is $b = 27$ .

Factor $27$ out of $27 u$ .

$- 6 x (x^{2} + 4) + 7 u^{2} + 27 (u) - 4 = 0$

Rewrite $27$ as $- 1$ plus $28$

$- 6 x (x^{2} + 4) + 7 u^{2} + (- 1 + 28) u - 4 = 0$

Apply the distributive property.

$- 6 x (x^{2} + 4) + 7 u^{2} - 1 u + 28 u - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 6 x (x^{2} + 4) + 7 u^{2} - 1 u + 28 u - 4 = 0$

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

$- 6 x (x^{2} + 4) + u (7 u - 1) + 4 (7 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 u - 1$ .

$- 6 x (x^{2} + 4) + (7 u - 1) (u + 4) = 0$

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

$- 6 x (x^{2} + 4) + (7 x^{2} - 1) (x^{2} + 4) = 0$

Factor $x^{2} + 4$ out of $- 6 x (x^{2} + 4) + (7 x^{2} - 1) (x^{2} + 4)$ .

Factor $x^{2} + 4$ out of $- 6 x (x^{2} + 4)$ .

$(x^{2} + 4) (- 6 x) + (7 x^{2} - 1) (x^{2} + 4) = 0$

Factor $x^{2} + 4$ out of $(7 x^{2} - 1) (x^{2} + 4)$ .

$(x^{2} + 4) (- 6 x) + (x^{2} + 4) (7 x^{2} - 1) = 0$

Factor $x^{2} + 4$ out of $(x^{2} + 4) (- 6 x) + (x^{2} + 4) (7 x^{2} - 1)$ .

$(x^{2} + 4) (- 6 x + 7 x^{2} - 1) = 0$

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

$(x^{2} + 4) (- 6 u + 7 u^{2} - 1) = 0$

Factor by grouping.

Reorder terms.

$(x^{2} + 4) (7 u^{2} - 6 u - 1) = 0$

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 $a \cdot c = 7 \cdot - 1 = - 7$ and whose sum is $b = - 6$ .

Factor $- 6$ out of $- 6 u$ .

$(x^{2} + 4) (7 u^{2} - 6 u - 1) = 0$

Rewrite $- 6$ as $1$ plus $- 7$

$(x^{2} + 4) (7 u^{2} + (1 - 7) u - 1) = 0$

Apply the distributive property.

$(x^{2} + 4) (7 u^{2} + 1 u - 7 u - 1) = 0$

Multiply $u$ by $1$ .

$(x^{2} + 4) (7 u^{2} + u - 7 u - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{2} + 4) ((7 u^{2} + u) - 7 u - 1) = 0$

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

$(x^{2} + 4) (u (7 u + 1) - (7 u + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 u + 1$ .

$(x^{2} + 4) ((7 u + 1) (u - 1)) = 0$

Factor.

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

$(x^{2} + 4) ((7 x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 4) (7 x + 1) (x - 1) = 0$

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

$x^{2} + 4 = 0$

$7 x + 1 = 0$

$x - 1 = 0$

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

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

$x^{2} + 4 = 0$

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

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

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

$x = \pm i \cdot \sqrt{4}$

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

$x = \pm i \cdot \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

Set $7 x + 1$ equal to $0$ and solve for $x$ .

Set $7 x + 1$ equal to $0$ .

$7 x + 1 = 0$

Solve $7 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$7 x = - 1$

Divide each term in $7 x = - 1$ by $7$ and simplify.

Divide each term in $7 x = - 1$ by $7$ .

$\frac{7 x}{7} = \frac{- 1}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 1}{7}$

Divide $x$ by $1$ .

$x = \frac{- 1}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{7}$

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.

$x = 1$

The final solution is all the values that make $(x^{2} + 4) (7 x + 1) (x - 1) = 0$ true.

$x = 2 i, - 2 i, - \frac{1}{7}, 1$

Step 3

Find the Roots (Zeros) f(x)=7x^4-6x^3+27x^2-24x-4

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