Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

Remember to place a  when there isn't a coefficient given.

We multiply what's below the line by  and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

 with remainder 

This can be rewritten as:

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

To divide synthetically, we begin by drawing a box. On the inside separated by spaces, we write the coefficients of the terms of our polynomial being divided. On the outside, we write the root that would satisfy our binomial , namely .  Leaving a space for another row of numbers, we then draw a line below our row of coefficients.

We then begin dividing by simply carrying our first coefficient (1) down below the line.

We then multiply this 1 by our divisor (3) and write the resulting product (3) below our next coefficient.

We then add the two numbers in that column and write the sum (5) below the line.

We then simply continue the process by multiplying this 5 by our divisor 3 and writing that product in the next column, adding it to the next coefficient, and continuing until we finish the columns.

We then need to translate our bottom row of numbers into the coefficients of our new quotient. Since the first column originally corresponded to our cubic term, it will now correspond to the quadratic term meaning that our 1 can be translated as .  Similarly, our second column transitions from quadratic to linear, making our 5 become .  Finally, our third column becomes the constant term, meaning 8 simply remains the constant 8.  Finally, our former constant column becomes the column for our remainder.  However, since we have a 0, we have no remainder and can disregard it.

Putting all of this together gives us a final answer of 

First, set up the synthetic division problem by lining up the coefficients. There are a couple of different strategies - for this one, we will put a -7 in the top corner and add the columns.

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The first step is to bring down the first 1. Then multiply what is below the line by the -7 in the box, write it below the next coefficient, and then add the columns:

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We can interpret this answer as meaning 

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by  and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficient.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

 with reminder 

This can be rewritten as:

To answer this question correctly, you need to know the basic method of polynomial division. It works like long division that you learned in elementary school, but instead of dividing numbers with multiple digits, you are dividing polynomials with multiple terms.

The sneaky trick in this question is that the expression  is presented with the remainder term first and the main quotient term second! This is the opposite of the standard form of such an expression. But it is still technically mathematically correct, since changing the order of the terms in an addition expression does not change its value. The SAT loves to try to confuse students with sneaky but mathematically valid tricks like this. Always be watching out for them!

So the first thing you should do is rearrange the expression in the question into its most natural standard form: . Now you can see that this expression is the result of a polynomial division operation, with 4 as the main quotient and 2 as the remainder.

Now that you see that 4 is the main quotient, you can recognize that it is the result of dividing  by , so they must be the first term of the numerator and the first term of the denominator (which we call the divisor). Therefore we can eliminate the wrong answer choices  and , and focus on the remaining answer choices with  in the numerator. This is actually the hardest part of solving this question. Many students are confused by the order of the terms and mistakenly take 2 as the main quotient and 4 as the remainder. This mistake can lead to the wrong answer choice  in particular.

It is important to understand that the remainder term is always a fractional term with the original divisor still present as the denominator of this fractional term as well. What we often think of as “the remainder”, such as 2 in this case, is actually the numerator of this remainder term.

We can also eliminate the wrong answer choice  because this division operation would produce the value 4 with no remainder. This leaves only the correct answer choice, . The divisor  goes into the numerator   four times, so the main quotient term is 4. We multiply  by 4 to get , the we subtract  to get 2. This is our remainder, giving us the remainder term . Thus we have correctly obtained both terms of the original expression in the question, .

To answer this question correctly, you need to know the basic method of polynomial division. It works like long division that you learned in elementary school, but instead of dividing numbers with multiple digits, you are dividing polynomials with multiple terms.

The first step of the polynomial division process in this question is fairly straightforward: The first term will be x because that is the result of dividing the first term of the numerator  by the first term of the denominator (the divisor) .

The first tricky part occurs when you multiply  by  to get , which you must write beneath  and subtract. Be very careful in performing the subtraction steps when solving polynomial division operations! In basic long division of numbers, all the digits are positive, so the subtraction steps are just like normal subtraction. But in polynomial division, the terms can be negative as well as positive, and it is trickier to perform subtraction correctly with negative terms. Therefore, it is a good idea to perform such subtraction steps by changing all the signs of the second polynomial and then adding. In this case, that means rather than trying to do the subtraction step , instead you should change all the signs of the second polynomial to  and then add: . Then you will get the correct result . Students who make a mistake in this step and get positive  as a result of the subtraction are likely to end up with one of the wrong answer choices  or .

The next step is to bring down the last term of the numerator -2 to make . Now you see that the divisor  goes into this polynomial -1 times, so the next term of the quotient is -1. The whole quotient is now .

Now you have to multiply  by -1 to get , which you must write beneath  and subtract. Again, as recommended above, now it is best to change all the sign of the second polynomial  to  and then add: . Then you will get the correct result -4. Students who make a mistake in this step may get 0 as a result of the subtraction and select the wrong answer choice .

Now that you have the quotient  and the remainder -4 , you can see that the correct answer choice is .

To answer this question correctly, you need to know the basic method of polynomial division. It works like the long division that you learned in elementary school, but instead of dividing numbers with multiple digits, you are dividing polynomials with multiple terms.

The first step of the polynomial division process in this question is fairly straightforward: The first term will be  because that is the result of dividing the first term of the numerator  by the first term of the denominator (the divisor) .

Next you multiply  by  to get , which you must write beneath  and subtract. The result of the subtraction is .

Now you have to understand the next step clearly: you can still do another polynomial division step and get another term in the quotient. The divisor  looks “bigger” than , but the important thing is that the first terms are the same, so the result of the division is positive 1. Now you have the whole quotient . Students who don’t realize they can get this second term of the quotient may select the wrong answer choice .

Next you have to multiply  by 1 to get , which you must write beneath  and subtract. This is a critical key step: You are really doing the subtraction , and so the correct result is -3. Students who make a mistake in this step may get positive 3 as a result and select the wrong answer choice .

Now that you have the quotient  and the remainder -3, you can see that the correct answer choice is .

This question is really testing your understanding of the concept of a factor of a polynomial. A factor of a polynomial is a divisor by which the polynomial is exactly divisible, with no remainder. This concept works in the same manner as it does with regular numbers: a factor of a number is a divisor by which the number is exactly divisible, with no remainder. For example, 5 is a factor of 30, because  exactly, with no remainder. But 5 is not a factor of 32, because  with a remainder of 2.

Here is an important concept to understand about factors of polynomials: If for example  is a factor of , then the value of  must be 0. This is actually the same connection that you find when you solve a quadratic equation by factoring it: If one of the factors is , then you know that  is a solution of the equation. Well, since you always make the other side of the quadratic equation 0 in order to solve it, what you have really found is that . Another way to describe this is that -1 is an x-intercept of the graph of the function .

Here is the key point about how this concept is related to this question: If you have no information about any values of the polynomial function  that are equal to 0, then you have no information about any factors of the polynomial function ! The only information this question gives us is that . Since we have no way to determine any values of  that are equal to 0, we have no way to determine what any of the factors of  are. By this fact alone, we already know that all three answer choices that mention a factor of  must be wrong!! By process of elimination, the correct answer choice must be "The remainder when  is divided by  is ."

While this problem is certainly doable with pure algebra, many complicated polynomial questions like this one are much more quickly solved by picking numbers. There are lots of  terms in this problem, but the problem specifically states that the equation must hold true for ALL values of  so you can turn a lot of complicated algebra into some rather quick arithmetic by picking an easy value for . You can't pick  because the only appearance of  is when it's multiplied by , but you can pick  leaving you with very small numbers to calculate. Doing so will leave you with:

And that simplifies quickly, to:

Which becomes:

And that means that  must equal .

You could of course, do the algebra here, but as a general rule when the polynomials involve several multiplicative terms like this on the SAT, the question usually gives you an opportunity to pick a simple number to make the work much easier.

When polynomials have this many terms on the SAT, and when you see a phrase like "for all values of ," it is usually faster and easier to pick a number for  to test which answer choices match the given information (that's typically when the question asks "which of the following expressions is equivalent to the above?") or, as in this case, to remove the commonly-occurring variable and solve for the less-common one. Here you have lots of  terms but only one  term, so since they say that this equation holds true for ALL VALUES of , you can pick an easy-to-use value like  and then use arithmetic, not multi-step algebra, to solve for .  If you plug in  the given equation becomes:

Which quickly simplifies to:

And that simplifies to:

So , meaning that .

Note that you could of course do the algebra, distributing all the multiplication in the given equation and then combining like terms. But on most polynomial questions with this much multiplication to do, the shortcut of picking a small number can save you lots of time and be less error-prone.