Solve left(x^2+5xright)^2-2(x^2+5x)=24

$${ \left( { x }^{ 2 } +5x \right) }^{ 2 } -2( { x }^{ 2 } +5x)=24$$

Solve for x

$$x=-6$$

$$x=1$$

$$x=-1$$

$$x=-4$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+5x\right)^{2}.

$$\left(x^{2}\right)^{2}+10x^{2}x+25x^{2}-2\left(x^{2}+5x\right)=24$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}+10x^{2}x+25x^{2}-2\left(x^{2}+5x\right)=24$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{4}+10x^{3}+25x^{2}-2\left(x^{2}+5x\right)=24$$

Use the distributive property to multiply -2 by x^{2}+5x.

$$x^{4}+10x^{3}+25x^{2}-2x^{2}-10x=24$$

Combine 25x^{2} and -2x^{2} to get 23x^{2}.

$$x^{4}+10x^{3}+23x^{2}-10x=24$$

Subtract 24 from both sides.

$$x^{4}+10x^{3}+23x^{2}-10x-24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+10x^{3}+23x^{2}-10x-24 by x-1 to get x^{3}+11x^{2}+34x+24. Solve the equation where the result equals to 0.

$$x^{3}+11x^{2}+34x+24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+11x^{2}+34x+24 by x+1 to get x^{2}+10x+24. Solve the equation where the result equals to 0.

$$x^{2}+10x+24=0$$

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, 10 for b, and 24 for c in the quadratic formula.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 24}}{2}$$

Do the calculations.

$$x=\frac{-10±2}{2}$$

Solve the equation x^{2}+10x+24=0 when ± is plus and when ± is minus.

$$x=-6 x=-4$$

List all found solutions.

$$x=1 x=-1 x=-6 x=-4$$