Solve x ^ 3 + p(x) ^ 2 + qx + 6 = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x^{3}+p(x)^{2}+qx+6=0$$

Solve for p

$p=-\frac{x^{3}+qx+6}{x^{2}}$
$x\neq 0$

Steps for Solving Linear Equation

Subtract $x^{3}$ from both sides. Anything subtracted from zero gives its negation.

$$px^{2}+qx+6=-x^{3}$$

Subtract $qx$ from both sides.

$$px^{2}+6=-x^{3}-qx$$

Subtract $6$ from both sides.

$$px^{2}=-x^{3}-qx-6$$

The equation is in standard form.

$$x^{2}p=-x^{3}-qx-6$$

Divide both sides by $x^{2}$.

$$\frac{x^{2}p}{x^{2}}=\frac{-x^{3}-qx-6}{x^{2}}$$

Dividing by $x^{2}$ undoes the multiplication by $x^{2}$.

$$p=\frac{-x^{3}-qx-6}{x^{2}}$$

Divide $-x^{3}-qx-6$ by $x^{2}$.

$$p=-\frac{qx+6}{x^{2}}-x$$

Solve for q

$q=-px-x^{2}-\frac{6}{x}$
$x\neq 0$

Steps for Solving Linear Equation

Subtract $x^{3}$ from both sides. Anything subtracted from zero gives its negation.

$$px^{2}+qx+6=-x^{3}$$

Subtract $px^{2}$ from both sides.

$$qx+6=-x^{3}-px^{2}$$

Subtract $6$ from both sides.

$$qx=-x^{3}-px^{2}-6$$

The equation is in standard form.

$$xq=-x^{3}-px^{2}-6$$

Divide both sides by $x$.

$$\frac{xq}{x}=\frac{-x^{3}-px^{2}-6}{x}$$

Dividing by $x$ undoes the multiplication by $x$.

$$q=\frac{-x^{3}-px^{2}-6}{x}$$

Divide $-x^{3}-px^{2}-6$ by $x$.

$$q=-px-x^{2}-\frac{6}{x}$$