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

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

Step 1

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

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

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

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

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

$x \cdot x^{2} - 3 x + 2 x^{4} - 7 x^{2} + 3 = 0$

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

$x \cdot x^{2} + x \cdot - 3 + 2 x^{4} - 7 x^{2} + 3 = 0$

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

$x (x^{2} - 3) + 2 x^{4} - 7 x^{2} + 3 = 0$

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

$x (x^{2} - 3) + 2 {(x^{2})}^{2} - 7 x^{2} + 3 = 0$

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

$x (x^{2} - 3) + 2 u^{2} - 7 u + 3 = 0$

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 u$ .

$x (x^{2} - 3) + 2 u^{2} - 7 u + 3 = 0$

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

$x (x^{2} - 3) + 2 u^{2} + (- 1 - 6) u + 3 = 0$

Apply the distributive property.

$x (x^{2} - 3) + 2 u^{2} - 1 u - 6 u + 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x (x^{2} - 3) + 2 u^{2} - 1 u - 6 u + 3 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

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

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

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

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

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

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

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

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

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

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

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

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

Factor by grouping.

Reorder terms.

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

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 $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

Rewrite $1$ as $- 1$ plus $2$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

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

Factor.

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

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

Remove unnecessary parentheses.

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

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

$x^{2} - 3 = 0$

$2 x - 1 = 0$

$x + 1 = 0$

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

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

$x^{2} - 3 = 0$

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

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

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.

$x = - 1$

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

$x = \sqrt{3}, - \sqrt{3}, \frac{1}{2}, - 1$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{3}, - \sqrt{3}, \frac{1}{2}, - 1$

Decimal Form:

$x = 1.73205080 \dots, - 1.73205080 \dots, 0.5, - 1$

Step 4

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

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