Solve (y^2+4y)^2+7(y^2+4y)+12=0

$$( y ^ { 2 } + 4 y ) ^ { 2 } + 7 ( y ^ { 2 } + 4 y ) + 12 = 0$$

Solve for y

$$y=-3$$

$$y=-1$$

$$y=-2$$

Solution Steps

Simplify.

$$y^{4}+8y^{3}+23y^{2}+28y+12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$y=-1$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{4}+8y^{3}+23y^{2}+28y+12 by y+1 to get y^{3}+7y^{2}+16y+12. Solve the equation where the result equals to 0.

$$y^{3}+7y^{2}+16y+12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$y=-2$$

By Factor theorem, y-k is a factor of the polynomial for each root k. Divide y^{3}+7y^{2}+16y+12 by y+2 to get y^{2}+5y+6. Solve the equation where the result equals to 0.

$$y^{2}+5y+6=0$$

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, 5 for b, and 6 for c in the quadratic formula.

$$y=\frac{-5±\sqrt{5^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$y=\frac{-5±1}{2}$$

Solve the equation y^{2}+5y+6=0 when ± is plus and when ± is minus.

$$y=-3 y=-2$$

List all found solutions.

$$y=-1 y=-2 y=-3$$