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Dividing Polynomials Using Synthetic Division 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). Set divisor = 0 and 3 2 1 x 6x 8x 2 solve. Put answer here. x3 x + 3 = 0 so x = - 3 -3 1 6 8 -2 up3these Bring number down below Addupthese line up - 3 firstAdd - 9theseAdd Multiply Multiply these these and and 2 +3 x - 1 This is the remainder 1 x 1 put answer put answer above line above line Put variables back in (one was of divided outthe in Sonext the Listanswer all coefficients is: (numbers in xfront x's) and in in next process so first number is one less 1 power 2 top. If a term is missing, constant along the putthan in a 0. column column original problem). x 3x 1 x3 Let's try another Synthetic Division Set divisor = 0 and solve. Put answer here. 4 1 0 x3 0 x 4 2 1 x 4x 6 x4 x - 4 = 0 so x = 4 0 -4 0 6 up48 Bring number down these below Add upthese line Add up these up 4 firstAdd 16theseAdd 192 Multiply Multiply Multiply these these and and 3 + 4 x2 + 12 x + 48 198 This is the these and 1 x put answer put answer remainder put answer above line above line Now put variables back in (remember one x was above lineanswer Sonext the List all coefficients is: (numbers in front of x's) and the198 in in next divided out 3in process2 so first number is one less in next constant along x the top. Don't forget the 0's48 for missing column 4 x 12 x column power than 3 original problem so x ). column terms. x4 Let's try a problem where we factor the polynomial completely given one of its factors. factor : x 2 4 x 8 x 25 x 50 3 -2 2 4 You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. 8 -25 -50 up Bring number down below Addupthese line up - 8 firstAdd 0theseAdd 50these Multiply Multiply these No remainder so x + 2 these and and 2 4 x + 0 x - 25 0 put IS a factor because it put answer answer above line divided in evenly above line Put variables back in (one x was divided outthe in Sonext the Listanswer all coefficients is the divisor (numbers times in thefront quotient: of x's) and in in next process sothe first number is one less power You could check this byIf a term constant along top. is missing, putthan in a 0. column column 2 original problem). multiplying them out and getting x 2 4 x 25 original polynomial 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