AS1.4: POLYNOMIAL LONG DIVISION One polynomial may be divided by another of lower degree by long division (similar to arithmetic long division). Example (x32+ 3 xx ++ 9) (x + 2) x + 2 x3 + 3x2 + x + 9 1. Write the question in long division form. 3 2. Begin with the x term. 2 x3 divided by x equals x2. x 2 Place x above the division bracket x+2 x32 + 3 xx ++ 9 as shown. subtract xx32+ 2 2 3. Multiply x + 2 by x . x2 xx2+=+22 x 32 x ( ) . Place xx32+ 2 below xx32+ 3 then subtract. x322+−+32 x x xx = 2 ( ) 4. “Bring down” the next term (x) as 2 x + x indicated by the arrow. +32 +++ Repeat this process with the result until xxx23x 9 there are no terms remaining. x32+↓2x 2 2 5. x divided by x equals x. x + x Multiply xx( +=+22) x2 x . xx2 +−1 ( xx22+−) ( x +2 x) =− x 32 x+23 x + xx ++ 9 6. Bring down the 9. xx32+2 ↓↓ 2 7. − x divided by x equals − 1. xx+↓ Multiply xx2 +↓2 −12( xx +) =−− 2 . −+x 9 (−+xx9) −−−( 2) = 11 −−x 2 The remainder is 11 +11 (x32+ 3 xx ++ 9) equals xx2 +−1, with remainder 11. (x + 2) AS 1.4 – Polynomial Long Division Page 1 of 4 June 2012 If the polynomial has an ‘xn ‘ term missing, add the term with a coefficient of zero. Example (2xx3 − 31 +÷) ( x − 1) Rewrite (2xx3 −+ 31) as (2xxx32+ 0 −+ 31) Divide using the method from the previous example. 2 22xx32÷= x 2xx+− 21 2 32 −= − 2xx( 12) x 2 x 32 x−12031 xxx + −+ 20xx32−=−−(22xx32) 2x2 2xx32− 2 ↓↓ 2 22xx÷= x 2xx2 −↓ 3 2xx( −= 12) x2 − 2 x 2 2 2 2xx−↓ 2 23xx−=−−22xx− x ( ) −+x 1 −÷xx =−1 −+x 1 −11( xx −) =−+ 1 0 −+xx1− ( + 10) = Remainder is 0 (2xx32− 31 +÷) ( x −= 1) ( 2 xx + 21 −) with remainder = 0 ∴(2xx32 − 3 += 1) ( x − 12)( xx + 2 − 1) See Exercise 1. Polynomial long division can be used help factorise cubic (and higher order) equations. AS 1.4 – Polynomial Long Division Page 2 of 4 June 2012 Solving Cubic Equations Factor Theorem If P(x) is a polynomial in x and P(a) = 0 then (x − a) is a factor of P(x) We can use the factor theorem to find one factor of a cubic function, and then use polynomial long division to find the remaining factor(s). (Sometimes it is possible to find all solutions by finding three values of x for which P(x) = 0 ). A cubic equation has a maximum of three distinct solutions. Example Solve the equation xxx32−5 −+= 2 24 0 Look for a value of x that makes P(x) = 0. Try x = 1 first, then work up through the factors of 24. Try x = 1 P(1) =18 x − 1 is not a factor Try x = 2 P(2) = 8 x − 2 is not a factor Try x = − 2 P(−2) = 0 P(−2) = 0 therefore x + 2 is a factor Next divide xxx32−5 −+ 2 24 by (x + 2), using the method of polynomial long division shown above. The remainder should be zero 2 xx−+7 12 x+2 xxx32 − 5 −+ 2 24 xx32+ 2 2 −−72xx −−7xx2 14 12x + 24 12x + 24 0 The factors of xxx32−5 −+ 2 24 are (x + 2) and (x2 − 7x + 12) The factors of xx2 −+7 12 are (x − 3) and (x − 4). xxx32−5 −+= 2 24 0 has solutions x = − 2, x = 3 and x = 4 AS 1.4 – Polynomial Long Division Page 3 of 4 June 2012 Exercises Exercise 1 Divide the first polynomial by the second. (a) 2xxx32− 6 ++ 5 2, x −2 (b) 3x32+ 13 xx +− 6 12, 3 x + 4, (c) 3xx3 +− 1, x +1 (d) xx3 +−2 3, x −1 (e) 2x43+ 5 xx ++ 3, 2 x + 1 (f) 2xx32+++ 5 x 12, 2 x + 3 Exercise 2 Solve the following equations. (a) xx32+7 + 11 x += 5 0 (b) 4xxx32+ 2 −= 20 (c) −x32 −3 xx ++= 30 (d) xx3 −7 −= 60 Answers Exercise 1 (a) 2xx2 −+ 21 remainder 4 (b) xx2 +−32 remainder -4 (c) 3xx2 −+ 34 remainder – 3 (d) xx2 ++3 remainder 0 (e) x32+21 xx −+ remainder 2 (f) xx2 −+4 remainder 0 Exercise 2 (a) x = − 5, x = - 1 (b) x = 0, x = 1, x = −1/2 (c) x = 1, x = − 3, x = −1 (d) x =−1, x = −2, x = 3 AS 1.4 – Polynomial Long Division Page 4 of 4 June 2012 .