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Algebraic Long Division: The Remainder and Factor Theorems

Learn how to divide polynomials using long division and find remainders & factors. Practice exercises included.

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Algebraic Long Division: The Remainder and Factor Theorems

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  1. Section 6.5 The Remainder and Factor Theorems

  2. Algebraic long division Divide 2x³ + 3x² - x + 1 by x + 2 x + 2 is the divisor 2x³ + 3x² - x + 1 is the dividend The quotient will be here.

  3. 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.

  4. Algebraic long division Now multiply 2x² by x + 2 and subtract

  5. Algebraic long division Bring down the next term, -x.

  6. Algebraic long division Now divide –x², the first term of –x² - x, by x, the first term of the divisor which gives –x.

  7. Algebraic long division Multiply –x by x + 2 and subtract

  8. Algebraic long division Bring down the next term, 1

  9. Algebraic long division Divide x, the first term of x + 1, by x, the first term of the divisor which gives 1

  10. Algebraic long division Multiply x + 2 by 1 and subtract

  11. Algebraic long division The quotient is 2x² - x + 1 The remainder is –1.

  12. Remainder • Since there is a remainder, (x+2) is not a factor of 2x3 + 3x2 –x + 1 • Remainder Theorem: If a polynomial f(x) is divided by x – k, the remainder is r = f(k). • Factor Theorem: If r = f(k) = 0, x – k is a factor of f(x).

  13. Methods to find the remainder • 1) Long division • 2) Simple substitution • 3) Synthetic division • You can use Synthetic division to divide a polynomial by an expression of the form x – k

  14. REMAINDER AND FACTOR THEOREMS f(x) = 2x2 – 3x + 4 Divide the polynomial by x – 2 Find f(2) 2 2 -3 4 f(2) = 2(2)2 – 3(2) + 4 2 4 f(2) = 8 – 6 + 4 2 1 6 f(2) = 6

  15. REMAINDER AND FACTOR THEOREMS Is x – 2 a factor of x3 – 3x2 – 4x + 12 2 1 -3 -4 12 Yes, it is a factor, since f(2) = 0. -12 -2 2 -6 0 1 -1 Can you find the two remaining factors?

  16. REMAINDER AND FACTOR THEOREMS (x + 3)( ? )( ? ) = x3 – x2 – 17x - 15 Find the two unknown ( ? ) quantities.

  17. REMAINDER AND FACTOR THEOREMS Rewrite: x3 – 3x2 – 4x + 12 as (x-2)(x2 –x -6) Factor x2 –x -6 x2 –x -6 = (x-3)(x+2) Factors are: -2, 2 and 3

  18. One of the zeros of x3 – 2x2 – 9x + 18 is x =2. What are the other zeros? 2 1 -2 -9 18 (x2 – 9)(x-2) =0 -18 0 2 x2 – 9 =(x+3)(x-3) -9 0 0 1 Factors are: x=2, x=-3, x=3

  19. Assignment Section 6.5: page 356 – 357 # 15 – 45 (÷6), 48 – 54 even

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