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Long division, synthetic division, remainder theorem and factor theorem


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Some chosen topics on Advanced Algebra, simplified

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Long division, synthetic division, remainder theorem and factor theorem

  1. 1. FACTOR THEOREM Long Division, Synthetic Division, Remainder Theorem &
  2. 2. INTRODUCTION In Advance Algebra, one of the most interesting topics are dividing Polynomials. Dividing is one of the most confusing operations among students. This presentation aims to teach us the 2 ways to divide polynomials and with 2 bonus topics: The Remainder Theorem and The Factor Theorem. This presentation is about to help most of us in dividing polynomials from a very complicated and confusing operation to an easy task.
  3. 3. TABLE OF CONTENTS Title Page………………………………….Slide 1 Introduction………………………………Sli de 2 Table of Contents………………………Slide 3 2 Ways to Divide Polynomials…….Slides 4- 8 Long Division Method……..........Slides 5- 6
  4. 4. FACTOR THEOREM Long Division Method Synthetic Division Remainder Theorem &  Is the usual way of dividing the polynomials that was introduced in Elementary Algebra STEPS: 1) Divide the first term of both the divisor and dividend. 2) Then, multiply the answer to the divisor. Put the product below the Dividend. Just be sure to align it properly. 3) Subtract and bring down the next digit or term. 4) Then, divide again the same way as before. 5) Repeat steps 2-4 until you get 0 or a remainder. q(x) – quotient r(x) - remainder
  5. 5. GIVEN: Use Long Division Method to divide 7x+3 to 14x2+20x+49 SOLUTION: 2x + 2 7x+3 14x2+20x+49 q(x) = 2x+2 -14x2+6x 14x+49 r(x) = 43 -14x+6 43 1) We divide 7x from 4x2 and we got 2x. 2) We multiply 2x to 7x+3. 3) We subtract, so we change signs (indicated by red marks). 4) The difference is 14x and we bring down 49. 5) We divide 7x from 14x and we got 2. 6) We then multiply 2 to 7x+3 and subtract it the same as in step 3. 7) Lastly, we got the remainder 43 and quotient 2x+2.
  6. 6.  Is another way of dividing polynomials by which we only use the constants among the given functions FACTOR THEOREM Long Division Synthetic Division Remainder Theorem & STEPS: 1) First, get the constants in the function as shown below: from x+4 8x2+6x+2 to 4 8 6 2 2) Bring down the first digit. 3) Multiply it to the divisor. Put the product below the next digit. 4) Subtract. 5) Repeat steps 2-4 until done. 6) Add xn-1 on the digits. Remember that the last digit would be the remainder. For instance, the answer for 12x2-15x+3 / x-2 is 12 9 21. Then, the final answer should be: q(x) = 12x+9 & r(x) = 21.
  7. 7. GIVEN: Use Synthetic division to divide x+2 from 4x4+2x3+x+5 SOLUTION: 2 4 2 0 1 5 q(x) = 4x3-6x2+12x-23 (-)8 -12 (-)24 -46 r(x) = 51 4 -6 12 -23 51 1) Since we don’t have a term with x2, we replaced it by 0 to complete the function. 2) We bring down 4. 3) We multiply it to 2 and we got 8. 4) We subtract, so we change signs. 5) We did the same to the next digits. 6) The final answer is 4x3-6x2+12x+23 and the remainder is 51. To prove that we are correct, we would use the Long Division method to solve the same problem….. .. .. .. 4x3-6x2+12x-23 x+2 4x4+2x3+x+5 q(x) = 4x3-6x2+12x-23 4x4+8x3 -6x3+x r(x) = 51 -6x3+12x2 12x2+x *We used 2 different 12x2+24x methods but we -23x+5 arrived w/ the same -23x-46 answer. 51
  8. 8. FACTOR THEOREM Long Division Synthetic Division Remainder Theorem &  It is simply a way in finding the remainder of a polynomial equation w/o finding its quotient STEPS: 1) Get the inverse of the constant of the divisor. This would be the value of x. 2) Substitute it to the dividend. 3) Solve. Perform the indicated operations. 4) The answer would be the remainder.
  9. 9. GIVEN: Consider the given in Synthetic division: SOLUTION: X = -2 4(-2)4+2(-2)3-2+5 4(16)+29(-8)+3 64-16+3 67-16 = 51 *Looking back to slide 8, the remainder were the same. 1) First, we get the inverse of constant of x+2 which is -2. It would be the value of x. 2) We substitute -2 to all x of the dividend. 3) We first solve the exponents and multiply. 4) Add the digits with the same sign. 5) Lastly, we add them and we got 51, the same remainder as we use Long and Synthetic Division Method.
  10. 10. FACTOR THEOREM Long Division, Synthetic Division, Remainder Theorem &  It is just a way of checking if the divisor is a factor of the dividend STEPS: 1) Multiply the divisor to the quotient. 2) Then, add the remainder (if any). 3) The answer should be the dividend. If not, then the divisor is not a factor of the dividend.
  11. 11. GIVEN: Consider the previous given (in Synthetic Division) SOLUTION: q(x) = 4x3-6x2+12x-23 divisor = x+2 r(x) = 51 dividend = 4x4+2x3+x+5 4x3-6x2+12x-23  q(x) x x+2  divisor 4x4-6x3+12x2-23x 8x3-12x2+24x-46 4x4+2x3+x-46 + 51  r(x) 4x4+2x3+x+5  dividend 1) We first multiply q(x) and the divisor. Then, we add r(x). 2) The answer is the dividend. Therefore, x+2 is a factor of 4x4+2x3+x+5.
  12. 12. GENERALIZATION There are 2 ways to divide polynomials: the Long Division Method and The Synthetic Division. The Long Division Method is the way of dividing polynomials that was taught in our 1st and 2nd year Algebra. The Synthetic Division is much easier and faster way in dividing polynomials. Remainder Theorem focuses on finding only the remainder even if not divided. Factor Theorem is just like checking operation in division of numbers.
  13. 13. Here is a table that contains different ways of dividing polynomials that includes the Remainder Theorem and The Factor Theorem. GIVEN: 4x4+2x3+x+5 / x+2 LONG DIVISION METHOD SYNTHETIC DIVISION REMAINDER THEOREM FACTOR THEOREM q(x) = 4x3-6x2+12x- 23 r(x) = 51 X = -2 4(-2)4+2(-2)3- 2+5 4(16)+29(-8)+3 64-16+3 67-16 = 51 4x3-6x2+12x-23 x x+2 4x4-6x3+12x2-23x 8x3-12x2+24x-46 4x4+2x3+x-46 + 51 4x4+2x3+x+5 Therefore, x+2 is a factor of 4x4+2x3+x+5. 2 4 2 0 1 5 (-)8 -12 (-)24 -46 4 -6 12 -23 51 4x3-6x2+12x-23 x+2 4x4+2x3+x+5 4x4+8x3 -6x3+x -6x3+12x2 12x2+x 12x2+24x -23x+5 -23x-46 51 q(x) = 4x3-6x2+12x-23 r(x) = 51
  14. 14. JOHN ROME R. ARANAS Creator SOURCES AND RESOURCES Math Notebook Myself New Century Mathematics By Phoenix Publishing House Ms. Charmaigne Marie Mahamis Google Wikepedia CREDITS