Solve 5x^3+6x^2+4x+14*Qx=7x^4-6x^3+4x

$$5 { x }^{ 3 } +6 { x }^{ 2 } +4x+14 \times Qx = 7 { x }^{ 4 } -6 { x }^{ 3 } +4x$$

Solve for Q

$$\left\{\begin{matrix}\\Q=-\frac{x\left(2-x\right)\left(7x+3\right)}{14}\text{, }&\text{unconditionally}\\Q\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 5x^{3} from both sides.

$$6x^{2}+4x+14Qx=7x^{4}-6x^{3}+4x-5x^{3}$$

Combine -6x^{3} and -5x^{3} to get -11x^{3}.

$$6x^{2}+4x+14Qx=7x^{4}-11x^{3}+4x$$

Subtract 6x^{2} from both sides.

$$4x+14Qx=7x^{4}-11x^{3}+4x-6x^{2}$$

Subtract 4x from both sides.

$$14Qx=7x^{4}-11x^{3}+4x-6x^{2}-4x$$

Combine 4x and -4x to get 0.

$$14Qx=7x^{4}-11x^{3}-6x^{2}$$

The equation is in standard form.

$$14xQ=7x^{4}-11x^{3}-6x^{2}$$

Divide both sides by 14x.

$$\frac{14xQ}{14x}=\frac{\left(x-2\right)\left(7x+3\right)x^{2}}{14x}$$

Dividing by 14x undoes the multiplication by 14x.

$$Q=\frac{\left(x-2\right)\left(7x+3\right)x^{2}}{14x}$$

Divide \left(-2+x\right)\left(3+7x\right)x^{2} by 14x.

$$Q=\frac{x\left(x-2\right)\left(7x+3\right)}{14}$$