Solve 4x+23/x^2+6x+9-3x+20/x^2-9-x+33/x^2-6x+9=0

$$\frac { 4 x + 23 } { x ^ { 2 } + 6 x + 9 } - \frac { 3 x + 20 } { x ^ { 2 } - 9 } - \frac { x + 33 } { x ^ { 2 } - 6 x + 9 } = 0$$

Solve for x

$$x=\frac{3}{10}=0.3$$

$$x=-5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}\left(x+3\right)^{2}, the least common multiple of x^{2}+6x+9,x^{2}-9,x^{2}-6x+9.

$$\left(x-3\right)^{2}\left(4x+23\right)-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\left(x^{2}-6x+9\right)\left(4x+23\right)-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}-6x+9 by 4x+23 and combine like terms.

$$4x^{3}-x^{2}-102x+207-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}-9 by 3x+20.

$$4x^{3}-x^{2}-102x+207-\left(3x^{3}+20x^{2}-27x-180\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

To find the opposite of 3x^{3}+20x^{2}-27x-180, find the opposite of each term.

$$4x^{3}-x^{2}-102x+207-3x^{3}-20x^{2}+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine 4x^{3} and -3x^{3} to get x^{3}.

$$x^{3}-x^{2}-102x+207-20x^{2}+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine -x^{2} and -20x^{2} to get -21x^{2}.

$$x^{3}-21x^{2}-102x+207+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine -102x and 27x to get -75x.

$$x^{3}-21x^{2}-75x+207+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Add 207 and 180 to get 387.

$$x^{3}-21x^{2}-75x+387-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{3}-21x^{2}-75x+387-\left(x^{2}+6x+9\right)\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}+6x+9 by x+33 and combine like terms.

$$x^{3}-21x^{2}-75x+387-\left(x^{3}+39x^{2}+207x+297\right)=0$$

To find the opposite of x^{3}+39x^{2}+207x+297, find the opposite of each term.

$$x^{3}-21x^{2}-75x+387-x^{3}-39x^{2}-207x-297=0$$

Combine x^{3} and -x^{3} to get 0.

$$-21x^{2}-75x+387-39x^{2}-207x-297=0$$

Combine -21x^{2} and -39x^{2} to get -60x^{2}.

$$-60x^{2}-75x+387-207x-297=0$$

Combine -75x and -207x to get -282x.

$$-60x^{2}-282x+387-297=0$$

Subtract 297 from 387 to get 90.

$$-60x^{2}-282x+90=0$$

Divide both sides by 6.

$$-10x^{2}-47x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -10x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-47 ab=-10\times 15=-150$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -47.

$$a=3 b=-50$$

Rewrite -10x^{2}-47x+15 as \left(-10x^{2}+3x\right)+\left(-50x+15\right).

$$\left(-10x^{2}+3x\right)+\left(-50x+15\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(10x-3\right)-5\left(10x-3\right)$$

Factor out common term 10x-3 by using distributive property.

$$\left(10x-3\right)\left(-x-5\right)$$

To find equation solutions, solve 10x-3=0 and -x-5=0.

$$x=\frac{3}{10} x=-5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}\left(x+3\right)^{2}, the least common multiple of x^{2}+6x+9,x^{2}-9,x^{2}-6x+9.

$$\left(x-3\right)^{2}\left(4x+23\right)-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\left(x^{2}-6x+9\right)\left(4x+23\right)-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}-6x+9 by 4x+23 and combine like terms.

$$4x^{3}-x^{2}-102x+207-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}-9 by 3x+20.

$$4x^{3}-x^{2}-102x+207-\left(3x^{3}+20x^{2}-27x-180\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

To find the opposite of 3x^{3}+20x^{2}-27x-180, find the opposite of each term.

$$4x^{3}-x^{2}-102x+207-3x^{3}-20x^{2}+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine 4x^{3} and -3x^{3} to get x^{3}.

$$x^{3}-x^{2}-102x+207-20x^{2}+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine -x^{2} and -20x^{2} to get -21x^{2}.

$$x^{3}-21x^{2}-102x+207+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine -102x and 27x to get -75x.

$$x^{3}-21x^{2}-75x+207+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Add 207 and 180 to get 387.

$$x^{3}-21x^{2}-75x+387-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{3}-21x^{2}-75x+387-\left(x^{2}+6x+9\right)\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}+6x+9 by x+33 and combine like terms.

$$x^{3}-21x^{2}-75x+387-\left(x^{3}+39x^{2}+207x+297\right)=0$$

To find the opposite of x^{3}+39x^{2}+207x+297, find the opposite of each term.

$$x^{3}-21x^{2}-75x+387-x^{3}-39x^{2}-207x-297=0$$

Combine x^{3} and -x^{3} to get 0.

$$-21x^{2}-75x+387-39x^{2}-207x-297=0$$

Combine -21x^{2} and -39x^{2} to get -60x^{2}.

$$-60x^{2}-75x+387-207x-297=0$$

Combine -75x and -207x to get -282x.

$$-60x^{2}-282x+387-297=0$$

Subtract 297 from 387 to get 90.

$$-60x^{2}-282x+90=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -60 for a, -282 for b, and 90 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-282\right)±\sqrt{\left(-282\right)^{2}-4\left(-60\right)\times 90}}{2\left(-60\right)}$$

Square -282.

$$x=\frac{-\left(-282\right)±\sqrt{79524-4\left(-60\right)\times 90}}{2\left(-60\right)}$$

Multiply -4 times -60.

$$x=\frac{-\left(-282\right)±\sqrt{79524+240\times 90}}{2\left(-60\right)}$$

Multiply 240 times 90.

$$x=\frac{-\left(-282\right)±\sqrt{79524+21600}}{2\left(-60\right)}$$

Add 79524 to 21600.

$$x=\frac{-\left(-282\right)±\sqrt{101124}}{2\left(-60\right)}$$

Take the square root of 101124.

$$x=\frac{-\left(-282\right)±318}{2\left(-60\right)}$$

The opposite of -282 is 282.

$$x=\frac{282±318}{2\left(-60\right)}$$

Multiply 2 times -60.

$$x=\frac{282±318}{-120}$$

Now solve the equation x=\frac{282±318}{-120} when ± is plus. Add 282 to 318.

$$x=\frac{600}{-120}$$

Divide 600 by -120.

$$x=-5$$

Now solve the equation x=\frac{282±318}{-120} when ± is minus. Subtract 318 from 282.

$$x=-\frac{36}{-120}$$

Reduce the fraction \frac{-36}{-120} to lowest terms by extracting and canceling out 12.

$$x=\frac{3}{10}$$

The equation is now solved.

$$x=-5 x=\frac{3}{10}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}\left(x+3\right)^{2}, the least common multiple of x^{2}+6x+9,x^{2}-9,x^{2}-6x+9.

$$\left(x-3\right)^{2}\left(4x+23\right)-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\left(x^{2}-6x+9\right)\left(4x+23\right)-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}-6x+9 by 4x+23 and combine like terms.

$$4x^{3}-x^{2}-102x+207-\left(x^{2}-9\right)\left(3x+20\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}-9 by 3x+20.

$$4x^{3}-x^{2}-102x+207-\left(3x^{3}+20x^{2}-27x-180\right)-\left(x+3\right)^{2}\left(x+33\right)=0$$

To find the opposite of 3x^{3}+20x^{2}-27x-180, find the opposite of each term.

$$4x^{3}-x^{2}-102x+207-3x^{3}-20x^{2}+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine 4x^{3} and -3x^{3} to get x^{3}.

$$x^{3}-x^{2}-102x+207-20x^{2}+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine -x^{2} and -20x^{2} to get -21x^{2}.

$$x^{3}-21x^{2}-102x+207+27x+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Combine -102x and 27x to get -75x.

$$x^{3}-21x^{2}-75x+207+180-\left(x+3\right)^{2}\left(x+33\right)=0$$

Add 207 and 180 to get 387.

$$x^{3}-21x^{2}-75x+387-\left(x+3\right)^{2}\left(x+33\right)=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{3}-21x^{2}-75x+387-\left(x^{2}+6x+9\right)\left(x+33\right)=0$$

Use the distributive property to multiply x^{2}+6x+9 by x+33 and combine like terms.

$$x^{3}-21x^{2}-75x+387-\left(x^{3}+39x^{2}+207x+297\right)=0$$

To find the opposite of x^{3}+39x^{2}+207x+297, find the opposite of each term.

$$x^{3}-21x^{2}-75x+387-x^{3}-39x^{2}-207x-297=0$$

Combine x^{3} and -x^{3} to get 0.

$$-21x^{2}-75x+387-39x^{2}-207x-297=0$$

Combine -21x^{2} and -39x^{2} to get -60x^{2}.

$$-60x^{2}-75x+387-207x-297=0$$

Combine -75x and -207x to get -282x.

$$-60x^{2}-282x+387-297=0$$

Subtract 297 from 387 to get 90.

$$-60x^{2}-282x+90=0$$

Subtract 90 from both sides. Anything subtracted from zero gives its negation.

$$-60x^{2}-282x=-90$$

Divide both sides by -60.

$$\frac{-60x^{2}-282x}{-60}=-\frac{90}{-60}$$

Dividing by -60 undoes the multiplication by -60.

$$x^{2}+\left(-\frac{282}{-60}\right)x=-\frac{90}{-60}$$

Reduce the fraction \frac{-282}{-60} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{47}{10}x=-\frac{90}{-60}$$

Reduce the fraction \frac{-90}{-60} to lowest terms by extracting and canceling out 30.

$$x^{2}+\frac{47}{10}x=\frac{3}{2}$$

Divide \frac{47}{10}, the coefficient of the x term, by 2 to get \frac{47}{20}. Then add the square of \frac{47}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{47}{10}x+\left(\frac{47}{20}\right)^{2}=\frac{3}{2}+\left(\frac{47}{20}\right)^{2}$$

Square \frac{47}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{47}{10}x+\frac{2209}{400}=\frac{3}{2}+\frac{2209}{400}$$

Add \frac{3}{2} to \frac{2209}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{47}{10}x+\frac{2209}{400}=\frac{2809}{400}$$

Factor x^{2}+\frac{47}{10}x+\frac{2209}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{47}{20}\right)^{2}=\frac{2809}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{47}{20}\right)^{2}}=\sqrt{\frac{2809}{400}}$$

Simplify.

$$x+\frac{47}{20}=\frac{53}{20} x+\frac{47}{20}=-\frac{53}{20}$$

Subtract \frac{47}{20} from both sides of the equation.

$$x=\frac{3}{10} x=-5$$