How does $\frac {(x^2 + 2x) - (a^2 + 2a)}{x-a}$ reduce to $x + a + 2$?

Question

With the given function:

$$f(x) = x^2 + 2x $$

I am trying to evaluate the following expression:

$$\frac {f(x) - f(a)}{x-a} $$

I've been informed that the solution is: $x + a + 2$, where $x \ne a$

But I don't know how to get there. Can someone help me understand the logic behind the solution and where I'm going wrong? I hope this question is appropriate. It's my first on the site.

I started by replacing the functions with their bodies respecting the given inputs.

$$\frac {(x^2 + 2x) - (a^2 + 2a)}{x-a}$$

Then, it appears to me that one could factor out the X's and the A's.

$$\frac {x(x + 2) - a(a + 2)}{x - a} $$

At this point I'm actually stumped. I can see there's an $x$ and an $a$ with subtraction between them, but I don't see how they're related.

But given the solution, I guess the $x-a$ in the denominator cancels out the $x-a$ in the numerator (somehow?). Then I'd be left with

$$(x+2)+(a+2)$$

Adding this up I get $x+a+4$. I'm not sure what problems like these are called, so I'm not sure what to search to find what I'm missing.

Thank you in advance for helping me.

Answer 1

$$\frac {x(x + 2) - a(a + 2)}{x - a} $$

You got so close:

$$\frac {x^2 - a^2+2x-2a}{x - a} $$

Then you just need to know $x^2-a^2=(x+a)(x-a)$.

More generally $x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+...+a^{n-1})$. This formula is used all over the place in maths. You can see it's true by considering what happens if you put $x=a$ into $x^n-a^n$.

Answer 2

$$\frac{(x^2+2x)-(a^2+2a)}{x-a} = \frac{(x^2-a^2)+2(x-a)}{x-a}=\frac{(x+a)(x-a)+2(x-a)}{x-a}=\frac{(x-a)[(x+a)+2]}{x-a} = x+a+2,\;\; \text{if $x\neq a$}$$

Answer 3

We have that

$$\frac {(x^2 + 2x) - (a^2 + 2a)}{x-a} =\frac {x^2 -a^2 + 2x-2a}{x-a}=\frac {(x+a)(x-a) + 2(x-a)}{x-a}=$$

$$=\frac {(x+a)(x-a) }{x-a}+\frac {2(x-a)}{x-a}=x+a+2$$

for $x\neq a$, the key point it to recognize that $x^2-a^2=(x+a)(x-a)$.

As an alternative we can proceed using that

$$x^2 +2x-a^2-2a=0 \implies x=\frac{-2\pm \sqrt{4-4(-a^2-2a)}}{2}=-1\pm(a+1)$$

and therefore, since $u=-a-2$ and $v=a$ are roots for the polynomial

$$x^2 +2x-a^2-2a=(x-u)(x-v)=(x+a+2)(x-a)$$

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Another way by long division

\begin{array}{rrr|ll} x^2 & +2x & -a^2-2a & x - a \\ -x^2 & +ax & & & x +a+2 \\ \hline & (a+2)x & -a^2-2a \\&-(a+2)x & a^2+2a & & & & \\ \hline & 0& 0 \end{array}