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Algebraic long division Divide 2x³ + 3x² - x + 1 by x + 2 x + 2 is the divisor x 2 2x 3 3x 2 x 1 The quotient will be here. 2x³ + 3x² - x + 1 is the dividend Algebraic long division First divide the first term of the dividend, 2x³, by x (the first term of the divisor). This gives 2x². This will be the first term of the quotient. 2x 2 x 2 2x 3 3x 2 x 1 Algebraic long division Now multiply 2x² by x + 2 and subtract 2x 2 x 2 2x 3 3x 2 x 1 2x 3 4x 2 x 2 Algebraic long division 2x 2 x 2 2x 3 3x 2 x 1 Bring down the next term, -x. 2x 3 4x 2 x 2 x Algebraic long division Now divide –x², 2x 2 x the first term of x 2 2x 3 3x 2 x 1 –x² - x, by x, the 3 2 2 x 4 x first term of the divisor x 2 x which gives –x. Algebraic long division 2x 2 x x 2 2x 3 3x 2 x 1 2x 3 4x 2 Multiply –x by x + 2 and subtract x 2 x x 2 2x x Algebraic long division 2x 2 x x 2 2x 3 3x 2 x 1 2x 3 4x 2 Bring down the next term, 1 x 2 x x 2 2x x 1 Algebraic long division 2x 2 x 1 x 2 2x 3 3x 2 x 1 Divide x, the first term of x + 1, by x, the first term of the divisor which gives 1 2x 3 4x 2 x 2 x x 2 2x x 1 Algebraic long division 2x 2 x 1 x 2 2x 3 3x 2 x 1 2x 3 4x 2 x 2 x x 2 2x Multiply x + 2 by 1 and subtract x 1 x 2 1 Algebraic long division 2x 2 x 1 x 2 2x 3 3x 2 x 1 The quotient is 2x² - x + 1 2x 3 4x 2 The remainder is –1. x 2 x x 2 2x x 1 x 2 1 You try one. x 3 2x 10x 12 2 Do the next two 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 6 x 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 power than 2 top. If a term is missing, put in a 0. constant along the column column original problem). 1 x 3x 1 x3 Let's try another Synthetic Division Set divisor = 0 and solve. Put answer here. 4 1 0 x3 0x 1 x 4x 6 4 2 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 the in in next divided out 3in process2so first number is one less in next constant along the top. Don't forget the 0's for missing column column power than original problem so x3). column terms. 198 x 4 x 12 x 48 x4 Let's try a problem where we factor the polynomial completely given one of its factors. 4 x 3 8 x 2 25 x 50 -2 4 factor : x 2 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 2 column original problem). multiplying them out and getting original polynomial x 24 x 25 You try one 3x 17 x 15 x 25 x 5 3 2 Try another one 10 x 50 x 800 x6 4 3 Do #’s 5 & 7 REMAINDER THEOREM When polynomial f(x) is divided by x – a, the remainder is f(a) f(x) = 2x2 – 3x + 4 Divide the polynomial by x – 2 2 2 -3 4 2 1 4 2 6 Find f(2) f(2) = 2(2)2 – 3(2) + 4 f(2) = 8 – 6 + 4 f(2) = 6 REMAINDER THEOREM When synthetic division is used to evaluate a function, it is called SYNTHETIC SUBSTITUTION. Try this one: Remember – Some terms are missing f(x) = 3x5 – 4x3 + 5x - 3 Find f(-3) Do #’s 8 & 9 FACTOR THEOREM The binomial x – a is a factor of the polynomial f(x) if and only if f(a) = 0. REMAINDER AND FACTOR THEOREMS Is x – 2 a factor of x3 – 3x2 – 4x + 12 2 1 -3 2 1 -1 -4 12 -2 -12 -6 0 Yes, it is a factor, since f(2) = 0. Can you find the two remaining factors? REMAINDER AND FACTOR THEOREMS (x + 3)( ? )( ? ) = x3 – x2 – 17x - 15 Find the two unknown ( ? ) quantities. Find all the zeros. One factor has been given. f ( x) x 9 x 23x 15; x 5 3 2 Find all the zeros. One factor has been given. f ( x) x x 14 x 24; x 3 3 2 Find all the zeros. One factor has been given. f ( x) x 4 3x3 13x 2 15x; x 3 You try #’s 10 - 13