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

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

Solve for y (complex solution)

$$y=\left(x-3\right)\left(x+2\right)$$

$$x\neq -3\text{ and }x\neq -1$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{\left(x^{2}-2x-3\right)\left(x^{2}+5x+6\right)}{x^{2}+4x+3}.

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

Cancel out \left(x+1\right)\left(x+3\right) in both numerator and denominator.

$$y=\left(x-3\right)\left(x+2\right)$$

Expand the expression.

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

Solve for y

$$y=\left(x-3\right)\left(x+2\right)$$

$$x\neq -1\text{ and }x\neq -3$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{\left(x^{2}-2x-3\right)\left(x^{2}+5x+6\right)}{x^{2}+4x+3}.

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

Cancel out \left(x+1\right)\left(x+3\right) in both numerator and denominator.

$$y=\left(x-3\right)\left(x+2\right)$$

Expand the expression.

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

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{\sqrt{4y+25}+1}{2}\text{, }&\text{unconditionally}\\x=\frac{-\sqrt{4y+25}+1}{2}\text{, }&y\neq 6\text{ and }y\neq -4\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{4y+25}+1}{2}\text{, }&y\geq -\frac{25}{4}\\x=\frac{-\sqrt{4y+25}+1}{2}\text{, }&y\neq -4\text{ and }y\geq -\frac{25}{4}\text{ and }y\neq 6\end{matrix}\right.$$