Algebra Examples

$x^{3} - 2 x^{2} + 5 x - 6 = 4 x^{2} + 14 x + 12$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

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

$x^{3} - 2 x^{2} + 5 x - 6 - 4 x^{2} = 14 x + 12$

Step 1.2

Subtract $14 x$ from both sides of the equation.

$x^{3} - 2 x^{2} + 5 x - 6 - 4 x^{2} - 14 x = 12$

Step 1.3

Subtract $12$ from both sides of the equation.

$x^{3} - 2 x^{2} + 5 x - 6 - 4 x^{2} - 14 x - 12 = 0$

Step 2

Simplify $x^{3} - 2 x^{2} + 5 x - 6 - 4 x^{2} - 14 x - 12$ .

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Step 2.1

Subtract $4 x^{2}$ from $- 2 x^{2}$ .

$x^{3} - 6 x^{2} + 5 x - 6 - 14 x - 12 = 0$

Step 2.2

Subtract $14 x$ from $5 x$ .

$x^{3} - 6 x^{2} - 9 x - 6 - 12 = 0$

Step 2.3

Subtract $12$ from $- 6$ .

$x^{3} - 6 x^{2} - 9 x - 18 = 0$

Step 3

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$

$q = \pm 1$

Step 4

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$