Solve (x^2+4x)^2-2(x^2+4x-3)-21=0

$$( x ^ { 2 } + 4 x ) ^ { 2 } - 2 ( x ^ { 2 } + 4 x - 3 ) - 21 = 0$$

Solve for x

$$x=-5$$

$$x=1$$

$$x=-1$$

$$x=-3$$

Solution Steps

Simplify.

$$x^{4}+8x^{3}+14x^{2}-8x-15=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±15,±5,±3,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+8x^{3}+14x^{2}-8x-15 by x-1 to get x^{3}+9x^{2}+23x+15. Solve the equation where the result equals to 0.

$$x^{3}+9x^{2}+23x+15=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±15,±5,±3,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+9x^{2}+23x+15 by x+1 to get x^{2}+8x+15. Solve the equation where the result equals to 0.

$$x^{2}+8x+15=0$$

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, 8 for b, and 15 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 15}}{2}$$

Do the calculations.

$$x=\frac{-8±2}{2}$$

Solve the equation x^{2}+8x+15=0 when ± is plus and when ± is minus.

$$x=-5 x=-3$$

List all found solutions.

$$x=1 x=-1 x=-5 x=-3$$