Precalculus Examples

$12 x^{4} + 6 x^{3} - \frac{9}{4} x^{2} - \frac{3}{4} x + \frac{3}{16}$

Step 1

Set $12 x^{4} + 6 x^{3} - \frac{9}{4} x^{2} - \frac{3}{4} x + \frac{3}{16}$ equal to $0$ .

$12 x^{4} + 6 x^{3} - \frac{9}{4} x^{2} - \frac{3}{4} x + \frac{3}{16} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Combine $x^{2}$ and $\frac{9}{4}$ .

$12 x^{4} + 6 x^{3} - \frac{x^{2} \cdot 9}{4} - \frac{3}{4} x + \frac{3}{16} = 0$

Step 2.1.2

Move $9$ to the left of $x^{2}$ .

$12 x^{4} + 6 x^{3} - \frac{9 \cdot x^{2}}{4} - \frac{3}{4} x + \frac{3}{16} = 0$

Step 2.1.3

Combine $x$ and $\frac{3}{4}$ .

$12 x^{4} + 6 x^{3} - \frac{9 x^{2}}{4} - \frac{x \cdot 3}{4} + \frac{3}{16} = 0$

Step 2.1.4

Move $3$ to the left of $x$ .

$12 x^{4} + 6 x^{3} - \frac{9 x^{2}}{4} - \frac{3 x}{4} + \frac{3}{16} = 0$

Step 2.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = - \frac{1}{2}, \frac{1}{4}$

Step 3