# Algebra - Polynomials Division

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 Author: randycapped ID: 267648 Filename: Algebra - Polynomials Division Updated: 2014-03-24 07:36:29 Tags: Polynomials Folders: Algebra Description: Dividing Polynomials. Show Answers:

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1. Dividing a polynomial by a monomial.

Step 1: Divide each term in the numerator (dividend) by the single term in the denominator (divisor).

Step 2: Simplify to get your quotient.

2. Dividing a polynomial by a monomial.

Step 1: Divide each term in the numerator (dividend) by the single term in the denominator (divisor).

Step 2: Simplify to get your quotient.

Step 3: Check.

3. Using Long Division.

Divide 2x3 – 9x2 + 15  by  2x – 5
• First off,note that there is a gap in the degrees of the terms of the dividend: the polynomial
• 2x3 – 9x2 + 15 has no x term. , so it is important that I leave space for a x-term column, just in case. I can create this space by turning the dividend into 2x3 – 9x2 + 0x + 15. This is a legitimate mathematical step: since I've only added zero, I haven't actually changed the value of anything.
• 1st: What do I need to multiply 2x by to get 2x3? That would be x2.

2nd: Multiply x2 by the divisor, 2x - 5.

• 3rd: Now, you will "add the opposite". Which is the same as subtracting.

• 4th: Bring down the next column and start the process again. Now, you will determine what you need to multiply 2x by to get -4x2. Which gives you -2x to place into the quotient.

• 5th: Continue the process until you end up with a remainder. This time, we have a remainder of -10.

Step 6: When you have a remainder, you need to remember to add the remainder to the polynomial part of the answer. Simply add the remainder as shown below.

4. What procedure would you use to divide this polynomial? And why?

You would use long division because divisor (x+2) has more than one divisor.
5. What procedure would you use to divide this polynomial? And why?
• Since the divisor has only one term, I'll divide 8x into into each term in the numerator.

6. Long Division. Zero Remainder. Easy Mode.

• 1) Consider both the leading terms of the dividend and divisor.
• 2) Divide the leading term of the dividend by the leading term of the divisor.
• 3) Place the partial quotient on top.

• 4) Now take the partial quotient you placed on top, 3x, and distribute into the divisor (2x + 4).
• 5) Position the product of (3x) and (2x+4) under the dividend.
• Make sure to align them by similar terms.

6) Perform subtraction by switching the signs of the bottom polynomial.

7) Proceed with regular addition vertically.

Notice that the first column from the left cancels each other out. Nice!

• 8) Carry down the next adjacent "unused" term of the dividend.

• 9) Next, look at the bottom polynomial, −14x−28, take its leading term which is −14x and divide it by the leading term of the divisor, 2x.
• 10) Again, place the partial quotient on top.

• 11) Use the partial quotient that you put up, −7, and distribute into the divisor. Seeing a pattern now?
• 12) Place the product of −7 and the divisor below as the last line of polynomial entry.

13)  Subtraction means you will switch the signs (in red).

14) Perform regular addition along the columns of similar terms.

• 15) This is great because the remainder is zero. It means the divisor is a factor of the dividend.
• The final answer is just the stuff on top of the division symbol.

7. Long Division.
Solution:  If you observe the dividend, it is missing some powers of variable x which are x3 and x2. I need to insert zero coefficients as placeholders for missing powers of the variable. This is a critical part to correctly apply the procedures in long division.

So I rewrite the original problem as . Now all x's are accounted for!

• 1) Focus on the leading terms inside and outside the division symbol.
• 2) Divide the first term of the dividend by the first term of the divisor.
• 3) Position the partial answer on top.

• 4) Use that partial answer placed on top, 3x2 to distribute into the divisor (x + 1).
• 5) Put the result under the dividend.

• 6) Subtract them together by making sure to switch the signs of the bottom terms before adding.

• 7) Carry down the next unused term of the dividend.

• 8) Looking at the bottom polynomial, −3x3 + 0x2, use the leading term −3x3 and divide it by the leading term of the divisor, x. Put the answer above the division symbol.

• 9) Multiply the answer you got previously, −3x3, and distribute into the divisor (x + 1).
• 10) Place the answer below then perform subtraction.

• 11) Bring down the next adjacent term of the dividend.

• 12) Go up again by dividing the leading term below by the leading term of the divisor.
• 13) Go down by distributing the answer in partial quotient into the divisor, followed by subtraction.

14) Carry down the last term of the dividend.

• 15) Go up again while performing division.

• 16) Go down again while performing multiplication.

17) Do the final subtraction, and we are done! Remainder is equal to 20.

• 18) The final answer in the form below is...
8. Long Division.

Solution: The dividend is obviously missing a lot of variable x. That means I need to insert zero coefficients in every missing powers of the variable.

I need to rewrite the problem this way to include all exponents of x: .

• Remember the main steps in long division:
• 1. When going up, we divide
• 2. When going down, we distribute
• 3. Subtract
• 4. Carry down
• 5. Repeat the process until done.

• Verify if the steps are being applied correctly in the example below.

So the final answer is: .
9. Long Division. Animated guide.

Solution:  We have a polynomial with five terms being divided by a trinomial. Both the dividend and divisor are in standard form, and all powers of the variable x are present. This is wonderful because we can now start solving it.

10. Long division. Missing "y" term in divisor.