Solve y^3+4y^2=9y+36

$$y ^ { 3 } + 4 y ^ { 2 } = 9 y + 36$$

Solve for y

$$y=-3$$

$$y=3$$

$$y=-4$$

Solution Steps

Subtract 9y from both sides.

$$y^{3}+4y^{2}-9y=36$$

Subtract 36 from both sides.

$$y^{3}+4y^{2}-9y-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$y=3$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}+4y^{2}-9y-36 by y-3 to get y^{2}+7y+12. Solve the equation where the result equals to 0.

$$y^{2}+7y+12=0$$

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 1 for a, 7 for b, and 12 for c in the quadratic formula.

$$y=\frac{-7±\sqrt{7^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$y=\frac{-7±1}{2}$$

Solve the equation y^{2}+7y+12=0 when ± is plus and when ± is minus.

$$y=-4 y=-3$$

List all found solutions.

$$y=3 y=-4 y=-3$$