Prove $\frac{(x^2-5x+2)}{(x+1)} = x-6$ has no solutions using contradiction

Question

I need help proving that this equation $$\frac{(x^2-5x+2)}{(x+1)} = x-6$$

has no solutions using contradiction.

I know that it has no solution because when you solve it you get $$\frac{8}{(x+1)} = 0$$

which is not true.

Answer 1

Let x=a be solution to $$\frac{(x^2-5x+2)}{(x+1)} = x-6$$

Then we have

$$\frac{(a^2-5a+2)}{(a+1)} = a-6$$

$${(a^2-5a+2)}=(a+1)(a-6)$$

$$a^2-5a+2 = a^2-5a-6$$

$$ 2=-6$$

This is a contradiction, so there is no solution.

Answer 2

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

$$(x^2-5x+2)=(x+1)(x+6)$$

$$x^2-5x+2=x^2+7x+6$$

$$ -12x =4$$

$$ x=-1/3$$

$ x=-1/3$ is indeed a solution.

Answer 3

Note that for all $x\neq -1$ we have:

$$x^2-5x+2 = (x-6)(x+1)+8\implies \frac{x^2-5x+2}{x+1} = x-6 + \frac 8{x+1}$$

and since $\frac 8{x+1}\neq 0$, your equation has no solutions.