Solve (x)=x^3-8/x^2-5x+6divx^2+2x+4/x-3

$$( x ) = \frac { x ^ { 3 } - 8 } { x ^ { 2 } - 5 x + 6 } \div \frac { x ^ { 2 } + 2 x + 4 } { x - 3 }$$

Solve for x

$$x=1$$

Solution Steps

Variable x cannot be equal to 3 since division by zero is not defined. Divide \frac{x^{3}-8}{x^{2}-5x+6} by \frac{x^{2}+2x+4}{x-3} by multiplying \frac{x^{3}-8}{x^{2}-5x+6} by the reciprocal of \frac{x^{2}+2x+4}{x-3}.

$$x=\frac{\left(x^{3}-8\right)\left(x-3\right)}{\left(x^{2}-5x+6\right)\left(x^{2}+2x+4\right)}$$

Use the distributive property to multiply x^{3}-8 by x-3.

$$x=\frac{x^{4}-3x^{3}-8x+24}{\left(x^{2}-5x+6\right)\left(x^{2}+2x+4\right)}$$

Use the distributive property to multiply x^{2}-5x+6 by x^{2}+2x+4 and combine like terms.

$$x=\frac{x^{4}-3x^{3}-8x+24}{x^{4}-3x^{3}-8x+24}$$

Subtract \frac{x^{4}-3x^{3}-8x+24}{x^{4}-3x^{3}-8x+24} from both sides.

$$x-\frac{x^{4}-3x^{3}-8x+24}{x^{4}-3x^{3}-8x+24}=0$$

Factor x^{4}-3x^{3}-8x+24.

$$x-\frac{x^{4}-3x^{3}-8x+24}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}.

$$\frac{x\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}-\frac{x^{4}-3x^{3}-8x+24}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}=0$$

Since \frac{x\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)} and \frac{x^{4}-3x^{3}-8x+24}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)-\left(x^{4}-3x^{3}-8x+24\right)}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}=0$$

Do the multiplications in x\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)-\left(x^{4}-3x^{3}-8x+24\right).

$$\frac{x^{5}-8x^{2}-3x^{4}+24x-x^{4}+3x^{3}+8x-24}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}=0$$

Combine like terms in x^{5}-8x^{2}-3x^{4}+24x-x^{4}+3x^{3}+8x-24.

$$\frac{x^{5}-8x^{2}-4x^{4}+32x+3x^{3}-24}{\left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right)}=0$$

Variable x cannot be equal to any of the values 2,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x-2\right)\left(x^{2}+2x+4\right).

$$x^{5}-8x^{2}-4x^{4}+32x+3x^{3}-24=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{5}-4x^{4}+3x^{3}-8x^{2}+32x-24=0$$

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^{5}-4x^{4}+3x^{3}-8x^{2}+32x-24 by x-1 to get x^{4}-3x^{3}-8x+24. Solve the equation where the result equals to 0.

$$x^{4}-3x^{3}-8x+24=0$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-3x^{3}-8x+24 by x-2 to get x^{3}-x^{2}-2x-12. Solve the equation where the result equals to 0.

$$x^{3}-x^{2}-2x-12=0$$

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.

$$x=3$$

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

$$x^{2}+2x+4=0$$

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, 2 for b, and 4 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-2±\sqrt{-12}}{2}$$

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

$$x\in \emptyset$$

Remove the values that the variable cannot be equal to.

$$x=1$$

List all found solutions.

$$x=1 x=2 x=3$$

Variable x cannot be equal to any of the values 2,3.

$$x=1$$