Solve x^2*(6+5/x+1)=4

$${ x }^{ 2 } \times (6+ \frac{ 5 }{ x+1 } )=4$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=-2$$

$$x=-\frac{1}{2}=-0.5$$

Solution Steps

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$x^{2}\left(6+\frac{5}{x+1}\right)\left(x+1\right)=4\left(x+1\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6 times \frac{x+1}{x+1}.

$$x^{2}\left(\frac{6\left(x+1\right)}{x+1}+\frac{5}{x+1}\right)\left(x+1\right)=4\left(x+1\right)$$

Since \frac{6\left(x+1\right)}{x+1} and \frac{5}{x+1} have the same denominator, add them by adding their numerators.

$$x^{2}\times \frac{6\left(x+1\right)+5}{x+1}\left(x+1\right)=4\left(x+1\right)$$

Do the multiplications in 6\left(x+1\right)+5.

$$x^{2}\times \frac{6x+6+5}{x+1}\left(x+1\right)=4\left(x+1\right)$$

Combine like terms in 6x+6+5.

$$x^{2}\times \frac{6x+11}{x+1}\left(x+1\right)=4\left(x+1\right)$$

Express x^{2}\times \frac{6x+11}{x+1} as a single fraction.

$$\frac{x^{2}\left(6x+11\right)}{x+1}\left(x+1\right)=4\left(x+1\right)$$

Express \frac{x^{2}\left(6x+11\right)}{x+1}\left(x+1\right) as a single fraction.

$$\frac{x^{2}\left(6x+11\right)\left(x+1\right)}{x+1}=4\left(x+1\right)$$

Use the distributive property to multiply 4 by x+1.

$$\frac{x^{2}\left(6x+11\right)\left(x+1\right)}{x+1}=4x+4$$

Use the distributive property to multiply x^{2} by 6x+11.

$$\frac{\left(6x^{3}+11x^{2}\right)\left(x+1\right)}{x+1}=4x+4$$

Use the distributive property to multiply 6x^{3}+11x^{2} by x+1 and combine like terms.

$$\frac{6x^{4}+17x^{3}+11x^{2}}{x+1}=4x+4$$

Subtract 4x from both sides.

$$\frac{6x^{4}+17x^{3}+11x^{2}}{x+1}-4x=4$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -4x times \frac{x+1}{x+1}.

$$\frac{6x^{4}+17x^{3}+11x^{2}}{x+1}+\frac{-4x\left(x+1\right)}{x+1}=4$$

Since \frac{6x^{4}+17x^{3}+11x^{2}}{x+1} and \frac{-4x\left(x+1\right)}{x+1} have the same denominator, add them by adding their numerators.

$$\frac{6x^{4}+17x^{3}+11x^{2}-4x\left(x+1\right)}{x+1}=4$$

Do the multiplications in 6x^{4}+17x^{3}+11x^{2}-4x\left(x+1\right).

$$\frac{6x^{4}+17x^{3}+11x^{2}-4x^{2}-4x}{x+1}=4$$

Combine like terms in 6x^{4}+17x^{3}+11x^{2}-4x^{2}-4x.

$$\frac{6x^{4}+17x^{3}+7x^{2}-4x}{x+1}=4$$

Subtract 4 from both sides.

$$\frac{6x^{4}+17x^{3}+7x^{2}-4x}{x+1}-4=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{x+1}{x+1}.

$$\frac{6x^{4}+17x^{3}+7x^{2}-4x}{x+1}-\frac{4\left(x+1\right)}{x+1}=0$$

Since \frac{6x^{4}+17x^{3}+7x^{2}-4x}{x+1} and \frac{4\left(x+1\right)}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6x^{4}+17x^{3}+7x^{2}-4x-4\left(x+1\right)}{x+1}=0$$

Do the multiplications in 6x^{4}+17x^{3}+7x^{2}-4x-4\left(x+1\right).

$$\frac{6x^{4}+17x^{3}+7x^{2}-4x-4x-4}{x+1}=0$$

Combine like terms in 6x^{4}+17x^{3}+7x^{2}-4x-4x-4.

$$\frac{6x^{4}+17x^{3}+7x^{2}-8x-4}{x+1}=0$$

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$6x^{4}+17x^{3}+7x^{2}-8x-4=0$$

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 6. List all candidates \frac{p}{q}.

$$±\frac{2}{3},±\frac{4}{3},±2,±4,±\frac{1}{3},±1,±\frac{1}{6},±\frac{1}{2}$$

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 6x^{4}+17x^{3}+7x^{2}-8x-4 by x+1 to get 6x^{3}+11x^{2}-4x-4. Solve the equation where the result equals to 0.

$$6x^{3}+11x^{2}-4x-4=0$$

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 6. List all candidates \frac{p}{q}.

$$±\frac{2}{3},±\frac{4}{3},±2,±4,±\frac{1}{3},±1,±\frac{1}{6},±\frac{1}{2}$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 6x^{3}+11x^{2}-4x-4 by x+2 to get 6x^{2}-x-2. Solve the equation where the result equals to 0.

$$6x^{2}-x-2=0$$

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 6 for a, -1 for b, and -2 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Do the calculations.

$$x=\frac{1±7}{12}$$

Solve the equation 6x^{2}-x-2=0 when ± is plus and when ± is minus.

$$x=-\frac{1}{2} x=\frac{2}{3}$$

Remove the values that the variable cannot be equal to.

$$x=\frac{2}{3}\text{ or }x=-\frac{1}{2}\text{ or }x=-2$$

List all found solutions.

$$x=-1 x=-2 x=-\frac{1}{2} x=\frac{2}{3}$$

Variable x cannot be equal to -1.

$$x=\frac{2}{3} x=-\frac{1}{2} x=-2$$