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Dividing Polynomials

Dividing Polynomials. divisor. Dividing by a Monomial. If the divisor only has one term, split the polynomial up into a fraction for each term. Now reduce each fraction. 3 x 3. 4 x 2. x. 2. 1. 1. 1. 1. Long Division. Now multiply by the divisor and put the answer below.

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Dividing Polynomials

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  1. Dividing Polynomials

  2. divisor Dividing by a Monomial If the divisor only has one term, split the polynomial up into a fraction for each term. Now reduce each fraction. 3x3 4x2 x 2 1 1 1 1

  3. Long Division Now multiply by the divisor and put the answer below. Bring down the next number or term 32 698 First divide 3 into 6 or x into x2 Now divide 3 into 5 or x into 11x So we found the answer to the problem x2 + 8x – 5  x – 3 or the problem written another way: If the divisor has more than one term, perform long division. You do the same steps with polynomial division as with integers. Let's do two problems, one with integers you know how to do and one with polynomials and copy the steps. Subtract (which changes the sign of each term in the polynomial) x + 11 2 1 Multiply and put below x - 3 x2 + 8x - 5 Remainder added here over divisor 64 x2 – 3x subtract 5 8 11x - 5 32 11x - 33 This is the remainder 26 28

  4. Write out with long division including 0y for missing term Divide y into y2 Divide y into -2y Let's Try Another One If any powers of terms are missing you should write them in with zeros in front to keep all of your columns straight. Subtract (which changes the sign of each term in the polynomial) y - 2 Multiply and put below Bring down the next term Multiply and put below y + 2 y2 + 0y + 8 Remainder added here over divisor y2 + 2y subtract -2y + 8 - 2y - 4 This is the remainder 12

  5. Set divisor = 0 and solve. Put answer here. x + 3 = 0 so x = - 3 Bring first number down below line Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Synthetic Division There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x). 1 - 3 1 6 8 -2 - 3 Add these up - 9 3 Add these up Add these up 1 x2 + x 3 - 1 1 This is the remainder Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.

  6. Set divisor = 0 and solve. Put answer here. x - 4 = 0 so x = 4 Bring first number down below line Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Let's try another Synthetic Division 0 x3 0 x 1 4 1 0 - 4 0 6 4 Add these up 16 48 192 Add these up Add these up Add these up 1 x3 + x2 + x + 4 12 48 198 This is the remainder Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x3). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms.

  7. Bring first number down below line Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Let's try a problem where we factor the polynomial completely given one of its factors. You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. - 2 4 8 -25 -50 - 8 Add these up 0 50 Add these up Add these up No remainder so x + 2 IS a factor because it divided in evenly 4 x2 + x 0 - 25 0 Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is the divisor times the quotient: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0. You could check this by multiplying them out and getting original polynomial

  8. Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au

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