Hi Mike, Can you work the solution for Test 7, Section 3, #13? Thanks!

Absolutely. First of all, this is a great one to plug in on. Equivalent means they’ll be the same for all values of x, so pick a value of x that’s easy to work with and get crackin’. I’m going to use x = 4, because that’ll make the denominator (x – 3) equal to 1.

The original expression:

    \begin{align*}&\dfrac{x^2-2x-5}{x-3}\\\\=&\dfrac{4^2-2(4)-5}{4-3}\\\\=&\dfrac{16-8-5}{1}\\\\=&3\end{align*}

Now, which answer choice gives you 3 when you plug in 4 for x? You can cross off A and B right away because they’re going to be negative—they start with x – 5! So try C and D:

    \begin{align*}\text{C)}\qquad &x+1-\dfrac{8}{x-3}\\\\=&4+1-\dfrac{8}{4-3}\\\\=&-3\\\\\text{D)}\qquad &x+1-\dfrac{2}{x-3}\\\\=&4+1-\dfrac{2}{4-3}\\\\=&3\end{align*}

Boom! D it is.

If you don’t want to plug in, you’re going to need to do polynomial division. (In other words, you should want to plug in!)

Polynomial long division is a real pain to render in text, so please forgive the handwriting.

Step 1: You’re always looking at the lead terms in long division. The leading term of the divisor, x, goes into the leading term of the dividend, x^2, exactly x times. Therefore, put an x at the top, and then subtract the product of x and x-3 from the dividend:

Step 2: x goes into x exactly 1 time, so put a + 1 on top and then subtract the product of x-3 and 1 from the x-5 we’re working with:

Step 3: Now you’re done, because x can’t go into –2. What’s on top is your quotient, and what’s left on bottom is your remainder.

The SAT will often show quotients and remainders as it does in the answer choices of this question: in the \text{QUOTIENT}+\dfrac{\text{REMAINDER}}{\text{DIVISOR}} form. In other words, x+1-\dfrac{2}{x-3} is the answer we’re looking for.