Dividing one polynomial by another, and writing what's left over as a fraction.
Last semester, your teacher handed back a graded test. Your score was 147 points out of a possible 4 problems — wait, that doesn't make sense. Points don't divide evenly into problems that way. But when you divide 147 by 4, you get 36 with a remainder of 3, which you write as 36\frac. Polynomial long division works exactly the same way. You divide, you get a quotient, and if something is left over, you write it as a fraction. The only difference is that your numbers are replaced by expressions in x.
The goal is this: given a polynomial p(x) divided by a polynomial b(x), find a quotient q(x) and remainder r(x) so that
\frac, = q(x) + \frac
The remainder r(x) must have a degree strictly less than the degree of b(x). That is the same rule as in arithmetic: when you divide by 4, the remainder must be less than 4. When you divide by a linear expression like x−2, the remainder must be a constant.
The procedure mirrors long division with integers. Divide the leading term of the dividend by the leading term of the divisor. Multiply the result by the entire divisor. Subtract. Bring down the next term. Repeat until the degree of what remains is less than the degree of the divisor.
Here is a clean first example.
Dividing x² + 5x + 6 by x + 2
xx2=x
Divide the leading term of the dividend by the leading term of the divisor. This is the first term of your quotient.
x⋅(x+2)=x2+2x
Multiply that first quotient term by the entire divisor.
(x2+5x+6)−(x2+2x)=3x+6
Subtract the whole thing. The x² terms cancel — that's the point. Now you have a simpler polynomial to work with.
x3x=3
Divide the new leading term by the leading term of the divisor. This is the second term of your quotient.
3⋅(x+2)=3x+6
Multiply again.
(3x+6)−(3x+6)=0
Subtract. Nothing is left. The remainder is zero, so x + 2 divides evenly into x² + 5x + 6.
x+2x2+5x+6=x+3✓
The quotient is x + 3. You can verify this by multiplying (x + 3)(x + 2) and confirming you get the original polynomial.
That example had no remainder. Most problems on the Regents will have one. Here is what that looks like.
Dividing 2x³ + 3x² - x + 5 by x + 2
x2x3=2x2
Divide leading terms. First term of the quotient is 2x².
2x2⋅(x+2)=2x3+4x2
Multiply 2x² by the entire divisor.
(2x3+3x2−x+5)−(2x3+4x2)=−x2−x+5
Subtract. Line up like terms carefully — a sign error here will throw off every step that follows.
x−x2=−x
Divide the new leading term. Second term of the quotient is -x.
−x⋅(x+2)=−x2−2x
Multiply -x by the entire divisor.
(−x2−x+5)−(−x2−2x)=x+5
Subtract. Subtracting a negative is addition — -x minus (-2x) gives +x.
xx=1
Divide the new leading term. Third term of the quotient is 1.
1⋅(x+2)=x+2
Multiply 1 by the entire divisor.
(x+5)−(x+2)=3
Subtract. What remains is 3, a constant. Its degree is less than the degree of x + 2, so we stop here. The remainder is 3.
x+22x3+3x2−x+5=2x2−x+1+x+23✓
Write the quotient, then add the remainder over the original divisor. This is the final answer form the Regents expects.
One situation that trips students up: a missing term. If you are dividing x3−8 by x−2, there are no x2 or x terms in the dividend. You must write placeholders — x3+0x2+0x−8 — or the columns will not line up and the subtraction step will produce garbage.
Interactive graph — scroll to zoom, drag to pan
The red curve is \frac and the blue curve is x2+2x+4. They overlap everywhere except at x=2, where the original expression is undefined. This confirms the division: x3−8 divided by x−2 gives x2+2x+4 with no remainder, for all x=2.
Practice Questions
x+5x2+7x+10
x−13x2−5x+1
x−3x3+0x2+0x−27
Regents Corner
On Part II and Part III of the Algebra II Regents, polynomial division problems require you to write the answer in the form q(x) + \frac. Leaving your answer as a long division diagram with a circled remainder earns no credit. The grader needs to see the complete expression. For a two-point problem, one point typically goes to the correct quotient and one to the correctly written remainder fraction.
Students subtract incorrectly when the term being subtracted is negative. In the step where you subtract −2x+2 from −2x+1, distributing the subtraction gives −2x+1+2x−2=−1. Students who write −2x+1−2x+2 and get −4x+3 have forgotten to flip every sign in the expression being subtracted, not just the first one.