How do you divide $\frac{x^{2} + 6 x + 5}{x + 5}$ $(x^2+6x+5) / (x+5)$ ?

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Answer 1 · 2015-06-29T19:06:35

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

Here, we can factor the numerator.

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

Now we can cancel $x + 5$ $x+5$ from numerator and denominator.

$(x^2 + 6x + 5)/(x+ 5) =(cancel((x+5))(x+1))/(cancel(x+5))$

So

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

How do you divide (x^2+6x+5) / (x+5)x2+6x+5x+5(x^2+6x+5) / (x+5)?