Algebra: Dividing Polynomials

Dividing Polynomials

There are two techniques you can use to calculate the quotient of two polynomials, one (which may feel a bit familiar) will work for all polynomial division problems but takes a while, whereas the other will work much faster, but only works in specific circumstances.

Long Division

The most reliable way to divide polynomials is the process of long division; even though the process is a bit cumbersome, it works for every division problem you'll see. It actually replicates the technique you learned in elementary school to divide whole numbers, in case you're wondering why it feels familiar.

Example 5: Calculate the quotient of (x³ + 5x² - 3x + 4) · (x² + 1).

Solution: Start by rewriting the problem in long division notation; the divisor (what you're dividing by) goes to the left, and the dividend (what you're dividing into) is written beneath the symbol. As you're writing the polynomials, make sure there are no missing exponents in either the divisor or dividend.

In this problem, the divisor has no x term; you don't want it to be missing, so you should write it in there with a coefficient of 0. (If you don't, things won't line up right.)

Look at the first term of the divisor, x², and the first term of the dividend, x³. Ask yourself, "What times x² will give me exactly x³?" The answer is x, and you should write that answer above the division symbol. In fact, you should write it directly above -3x, since it and the number you just came up with (x) are like terms.

Multiply that x and the divisor together, and write the result below the dividend, so that the like terms line up; draw a horizontal line beneath the product.

Now multiply everything in that bottom line by -1 and then combine the result with the like terms directly above. Write the result below the horizontal line.

Drop down the next term in the dividend polynomial, in this case a positive 4, and repeat the above process, beginning with the question, "What times x² (the first term in the divisor) will give me exactly 5x² (the first term in the subtraction problem you just finished); the answer is 5. Write that constant above the division symbol (right above 4, its like term), multiply it times the divisor, change each of the terms in the product to its opposite, and combine the like terms.

If there had been additional terms in the dividend, you'd drop them down one at a time and repeat the whole process again, but since there are no more terms to drop, you're finished. The quotient is the quantity above the division symbol, x + 5, and the remainder is -4x - 1.

You should write your answer as the quotient plus the fraction whose numerator is the remainder and whose denominator is the original divisor, like so:

• x + 5 + ^{-4x - 1}„_{x² + 1}

You can check your answer by multiplying the quotient (x + 5) times the original divisor (x² + 1) and then adding the remainder.

• (quotient)(divisor)+(remainder)
• =(x+5)(x²+1)+(-4x-1)
• =x³ + x + 5x² + 5 - 4x-1
• =x³+5x² + x- 4x+ 5 - 1
• =x³ +5x²-3x+4

If you did everything correctly, you should get the original dividend; that's exactly what happened here, so you can bask in the glow of your own mathematical greatness, secure in the knowledge that you rule.

Synthetic Division

If you're trying to divide a polynomial by a linear binomial (in the form "x - c" where c could be any real number), then the best way to do it is through synthetic division.

For example, the division problem (x³ - 2x² + 3 - 4) · (x + 3) would be a good candidate for synthetic division; the divisor is technically in the form x - c, because if you set c = -3, then x - c = x - (-3) = x + 3. However, the problem (x³ - 2x² + 3x - 4) · (x² + 3) would not be a good candidate, since the divisor is not linear.

Synthetic division is much simpler than long division because all you use are the coefficients of the polynomials. I don't know why it's called "synthetic" division€”it's not unnatural, filled with preservatives, or fake. It's never been to the plastic surgeon to get a nip, a tuck, an implant, or a reduction, so the name puzzles me. Like long division, the best way to learn the process is through an example, so I'll get right to it.

Example 6: Calculate the quotient of (2x³ - x + 4) · (x + 3).

Solution: First check to see if there are any missing powers of x in the dividend; notice that there's no x² power in there, so insert it with a coefficient of 0 (just like you did in long division) to get a dividend of 2x³ + 0x² - x + 4. Now list those coefficients in descending order of their exponents.

• 2 0 -1 4

To the left of that list, write the opposite of the constant in the divisor. In this problem, the divisor is x + 3, so the opposite of its constant will be -3. It's separated from the rest of the coefficients with something that looks like a half-box. Leave some space beneath that row and draw a horizontal line.

The setup is all finished, and it's time to get started. Take the leading coefficient (2) and drop it below the horizontal line.

Multiply the number in the half-box (-3) times the number below the line (2) and write the result (-6) below the next coefficient (0).

Combine the numbers in the second column (0 - 6 = -6) and write the result directly below the numbers you just combined.

Repeat the process two more times, each time multiplying the number in the box by the new number below the line, writing the result in the next column, adding the numbers in that column together, and writing the result below the line again.

The answer will be each of the numbers below the line with decreasing powers of x next to them; start with the x power one less than the degree of the dividend. Therefore, the first power of x in your answer for this problem should be 2. In case you're wondering, the rightmost number below the line is the remainder, which is written just like it was in long division.

• 2x² - 6x + 17 + ^-47„_{x + 3}

You can check this the same way you did long division: (2x² - 6x + 17)(x + 3) - 47 should equal 2x³ - x + 4, and it does.

Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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