Solve 12(x+1)(x+3)/(x-1)^5=0

$$12 \frac { ( x + 1 ) ( x + 3 ) } { ( x - 1 ) ^ { 5 } } = 0$$

Solve for x

$$x=-3$$

$$x=-1$$

Steps Using Factoring

Divide both sides by 12. Zero divided by any non-zero number gives zero.

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

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

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

$$x^{2}+4x+3=0$$

To solve the equation, factor x^{2}+4x+3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve x+1=0 and x+3=0.

$$x=-1 x=-3$$

Steps Using Factoring By Grouping

Divide both sides by 12. Zero divided by any non-zero number gives zero.

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

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

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

$$x^{2}+4x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 3=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=3$$

Rewrite x^{2}+4x+3 as \left(x^{2}+x\right)+\left(3x+3\right).

$$\left(x^{2}+x\right)+\left(3x+3\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x+1\right)+3\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve x+1=0 and x+3=0.

$$x=-1 x=-3$$

Steps Using the Quadratic Formula

Divide both sides by 12. Zero divided by any non-zero number gives zero.

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

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

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

$$x^{2}+4x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12}}{2}$$

Add 16 to -12.

$$x=\frac{-4±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-4±2}{2}$$

Now solve the equation x=\frac{-4±2}{2} when ± is plus. Add -4 to 2.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-4±2}{2} when ± is minus. Subtract 2 from -4.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=-1 x=-3$$

Steps for Completing the Square

Divide both sides by 12. Zero divided by any non-zero number gives zero.

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

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)^{5}.

$$\left(x+1\right)\left(x+3\right)=0$$

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

$$x^{2}+4x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+4x=-3$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=-3+2^{2}$$

Square 2.

$$x^{2}+4x+4=-3+4$$

Add -3 to 4.

$$x^{2}+4x+4=1$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+2=1 x+2=-1$$

Subtract 2 from both sides of the equation.

$$x=-1 x=-3$$