Solve 36y^3-4=y-144y^2

$$36 y ^ { 3 } - 4 = y - 144 y ^ { 2 }$$

Solve for y

$$y=-\frac{1}{6}\approx -0.166666667$$

$$y=\frac{1}{6}\approx 0.166666667$$

$$y=-4$$

Solution Steps

Subtract y from both sides.

$$36y^{3}-4-y=-144y^{2}$$

Add 144y^{2} to both sides.

$$36y^{3}-4-y+144y^{2}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$36y^{3}+144y^{2}-y-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 36. List all candidates \frac{p}{q}.

$$±\frac{1}{9},±\frac{2}{9},±\frac{1}{3},±\frac{4}{9},±\frac{2}{3},±1,±\frac{4}{3},±2,±4,±\frac{1}{18},±\frac{1}{6},±\frac{1}{2},±\frac{1}{36},±\frac{1}{12},±\frac{1}{4}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$y=-4$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 36y^{3}+144y^{2}-y-4 by y+4 to get 36y^{2}-1. Solve the equation where the result equals to 0.

$$36y^{2}-1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 36 for a, 0 for b, and -1 for c in the quadratic formula.

$$y=\frac{0±\sqrt{0^{2}-4\times 36\left(-1\right)}}{2\times 36}$$

Do the calculations.

$$y=\frac{0±12}{72}$$

Solve the equation 36y^{2}-1=0 when ± is plus and when ± is minus.

$$y=-\frac{1}{6} y=\frac{1}{6}$$

List all found solutions.

$$y=-4 y=-\frac{1}{6} y=\frac{1}{6}$$