Solve (x)=(2x+1)(x-5)/(2x-1)(x+3)?

$$( x ) = \frac { ( 2 x + 1 ) ( x - 5 ) } { ( 2 x - 1 ) ( x + 3 ) } ?$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{39}i-1}{4}\approx -0.25-1.5612495i$$

$$x=-1$$

$$x=\frac{-1+\sqrt{39}i}{4}\approx -0.25+1.5612495i$$

Solution Steps

Use the distributive property to multiply 2x+1 by x-5 and combine like terms.

$$x=\frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)}$$

Use the distributive property to multiply 2x-1 by x+3 and combine like terms.

$$x=\frac{2x^{2}-9x-5}{2x^{2}+5x-3}$$

Subtract \frac{2x^{2}-9x-5}{2x^{2}+5x-3} from both sides.

$$x-\frac{2x^{2}-9x-5}{2x^{2}+5x-3}=0$$

Factor 2x^{2}+5x-3.

$$x-\frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{\left(2x-1\right)\left(x+3\right)}{\left(2x-1\right)\left(x+3\right)}.

$$\frac{x\left(2x-1\right)\left(x+3\right)}{\left(2x-1\right)\left(x+3\right)}-\frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)}=0$$

Since \frac{x\left(2x-1\right)\left(x+3\right)}{\left(2x-1\right)\left(x+3\right)} and \frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(2x-1\right)\left(x+3\right)-\left(2x^{2}-9x-5\right)}{\left(2x-1\right)\left(x+3\right)}=0$$

Do the multiplications in x\left(2x-1\right)\left(x+3\right)-\left(2x^{2}-9x-5\right).

$$\frac{2x^{3}+6x^{2}-x^{2}-3x-2x^{2}+9x+5}{\left(2x-1\right)\left(x+3\right)}=0$$

Combine like terms in 2x^{3}+6x^{2}-x^{2}-3x-2x^{2}+9x+5.

$$\frac{2x^{3}+3x^{2}+6x+5}{\left(2x-1\right)\left(x+3\right)}=0$$

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(x+3\right).

$$2x^{3}+3x^{2}+6x+5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

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 2x^{3}+3x^{2}+6x+5 by x+1 to get 2x^{2}+x+5. Solve the equation where the result equals to 0.

$$2x^{2}+x+5=0$$

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 2 for a, 1 for b, and 5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-39}}{4}$$

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

$$x=\frac{-\sqrt{39}i-1}{4} x=\frac{-1+\sqrt{39}i}{4}$$

List all found solutions.

$$x=-1 x=\frac{-\sqrt{39}i-1}{4} x=\frac{-1+\sqrt{39}i}{4}$$

Solve for x

$$x=-1$$

Solution Steps

Use the distributive property to multiply 2x+1 by x-5 and combine like terms.

$$x=\frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)}$$

Use the distributive property to multiply 2x-1 by x+3 and combine like terms.

$$x=\frac{2x^{2}-9x-5}{2x^{2}+5x-3}$$

Subtract \frac{2x^{2}-9x-5}{2x^{2}+5x-3} from both sides.

$$x-\frac{2x^{2}-9x-5}{2x^{2}+5x-3}=0$$

Factor 2x^{2}+5x-3.

$$x-\frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{\left(2x-1\right)\left(x+3\right)}{\left(2x-1\right)\left(x+3\right)}.

$$\frac{x\left(2x-1\right)\left(x+3\right)}{\left(2x-1\right)\left(x+3\right)}-\frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)}=0$$

Since \frac{x\left(2x-1\right)\left(x+3\right)}{\left(2x-1\right)\left(x+3\right)} and \frac{2x^{2}-9x-5}{\left(2x-1\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(2x-1\right)\left(x+3\right)-\left(2x^{2}-9x-5\right)}{\left(2x-1\right)\left(x+3\right)}=0$$

Do the multiplications in x\left(2x-1\right)\left(x+3\right)-\left(2x^{2}-9x-5\right).

$$\frac{2x^{3}+6x^{2}-x^{2}-3x-2x^{2}+9x+5}{\left(2x-1\right)\left(x+3\right)}=0$$

Combine like terms in 2x^{3}+6x^{2}-x^{2}-3x-2x^{2}+9x+5.

$$\frac{2x^{3}+3x^{2}+6x+5}{\left(2x-1\right)\left(x+3\right)}=0$$

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(x+3\right).

$$2x^{3}+3x^{2}+6x+5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{5}{2},±5,±\frac{1}{2},±1$$

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 2x^{3}+3x^{2}+6x+5 by x+1 to get 2x^{2}+x+5. Solve the equation where the result equals to 0.

$$2x^{2}+x+5=0$$

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 2 for a, 1 for b, and 5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-39}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-1$$