Solve by Factoring 3x^4+11x^3=4x^2

$3 x^{4} + 11 x^{3} = 4 x^{2}$

Step 1

Subtract $4 x^{2}$ from both sides of the equation.

$3 x^{4} + 11 x^{3} - 4 x^{2} = 0$

Step 2

Factor $x^{2}$ out of $3 x^{4} + 11 x^{3} - 4 x^{2}$ .

Factor $x^{2}$ out of $3 x^{4}$ .

$x^{2} (3 x^{2}) + 11 x^{3} - 4 x^{2} = 0$

Factor $x^{2}$ out of $11 x^{3}$ .

$x^{2} (3 x^{2}) + x^{2} (11 x) - 4 x^{2} = 0$

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

$x^{2} (3 x^{2}) + x^{2} (11 x) + x^{2} \cdot - 4 = 0$

Factor $x^{2}$ out of $x^{2} (3 x^{2}) + x^{2} (11 x)$ .

$x^{2} (3 x^{2} + 11 x) + x^{2} \cdot - 4 = 0$

Factor $x^{2}$ out of $x^{2} (3 x^{2} + 11 x) + x^{2} \cdot - 4$ .

$x^{2} (3 x^{2} + 11 x - 4) = 0$

Step 3

Factor.

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 $b = 11$ .

Factor $11$ out of $11 x$ .

$x^{2} (3 x^{2} + 11 (x) - 4) = 0$

Rewrite $11$ as $- 1$ plus $12$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

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

Remove unnecessary parentheses.

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

Step 4

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

$x^{2} = 0$

$3 x - 1 = 0$

$x + 4 = 0$

Step 5

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

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

$x^{2} = 0$

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

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 6

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 $x$ .

Add $1$ to 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 $1$ .

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

Step 7

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.

$x = - 4$

Step 8

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

$x = 0, \frac{1}{3}, - 4$

Solve by Factoring 3x^4+11x^3=4x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.