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1. Use synthetic substitution to evaluate f ( x ) = x 3 + x 2 – 3 x – 10 when x = 2 .

2 1 1 – 3 -10. Warm Up #1. 1. Use synthetic substitution to evaluate f ( x ) = x 3 + x 2 – 3 x – 10 when x = 2. 2. 6. 6. 1. 3. 3. -4. – 4. ANSWER. Check HW 5.4 multiples of 3. EXAMPLE 1. Use polynomial long division.

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1. Use synthetic substitution to evaluate f ( x ) = x 3 + x 2 – 3 x – 10 when x = 2 .

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  1. 2 1 1 –3 -10 Warm Up #1 1.Use synthetic substitution to evaluate f(x)=x3+x2–3x–10 when x = 2. 2 6 6 1 3 3 -4 –4 ANSWER

  2. Check HW 5.4multiples of 3

  3. EXAMPLE 1 Use polynomial long division Divide f (x) = 3x4 – 5x3 + 4x – 6 byx2 – 3x + 5. SOLUTION Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 in the dividend. At each stage, divide the term with the highest power in what is left of the dividend by the first term of the divisor. This gives the next term of the quotient.

  4. 3x4 – 5x3 + 4x – 6 –25x + 9 quotient ANSWER = 3x2 + 4x – 3 + ) x2 – 3x + 5 x2 – 3x + 5 x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6 -(3x4 – 9x3 + 15x2) -(4x3– 12x2 + 20x) -(–3x2 + 9x – 15) remainder EXAMPLE 1 Use polynomial long division 3x2 + 4x – 3 4x3 – 15x2 + 4x –3x2 – 16x – 6 –25x + 9

  5. quotient ) x – 2 x3 + 5x2 – 7x + 2 -(x3 – 2x2) -(7x2– 14x) -(7x – 14) remainder x3 + 5x2– 7x +2 16 ANSWER = x2 + 7x + 7 + x – 2 x – 2 EXAMPLE 2 Use polynomial long division with a linear divisor Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2. + 7x + 7 x2 7x2 – 7x 7x + 2 16

  6. –18x + 7 ANSWER (2x2 – 3x + 8) + x2 + 2x – 1 –30 ANSWER (x2 – 3x + 10) + x + 2 for Examples 1 and 2 GUIDED PRACTICE Divide using polynomial long division. 1. (2x4 + x3 + x – 1) (x2 + 2x – 1) 2. (x3–x2 + 4x – 10)  (x + 2)

  7. –30 ANSWER (x2 – 3x + 10) + x + 2 for Examples 1 and 2 GUIDED PRACTICE Divide using polynomial long division. 2. (x3–x2 + 4x – 10)  (x + 2)

  8. –3 2 1 –8 5 EXAMPLE 3 Use synthetic division Dividef(x)= 2x3 + x2– 8x + 5byx +3using synthetic division. SOLUTION -6 15 -21 2 -5 7 -16

  9. –2 3 -4 -28 -16 EXAMPLE 4 Factor a polynomial Factorf (x) = 3x3– 4x2– 28x – 16completely given that x + 2is a factor. SOLUTION -6 20 16 3 -10 -8 0

  10. EXAMPLE 4 Factor a polynomial Use the result to write f (x) as a product of two factors and then factor completely. f (x) = 3x3– 4x2– 28x – 16 Write original polynomial. = (x + 2)(3x2– 10x – 8) Write as a product of two factors. = (x + 2)(3x + 2)(x – 4) Factor trinomial.

  11. 11 ANSWER x2+ x – 4 + –3 1 4 -1 -1 x + 3 for Examples 3 and 4 GUIDED PRACTICE Divide using synthetic division. 3. (x3 + 4x2–x – 1)  (x + 3) -3 -3 12 1 1 -4 11

  12. 4 1 -6 5 12 for Examples 3 and 4 GUIDED PRACTICE Factor the polynomial completely given that x –4 is a factor. 5. f (x) = x3– 6x2 + 5x + 12 (x – 4)(x –3)(x + 1) 4 -8 -12 1 -2 -3 0

  13. –2 1 2 -9 -18 for Examples 5 and 6 GUIDED PRACTICE Find the other zeros of fgiven that f (–2) = 0. 7. f (x) = x3 + 2x2– 9x – 18 -2 0 18 Zeros are -2, 3, -3 1 0 -9 0

  14. 3 1 –2 –23 60 3 3 –60 1 1 –20 0 EXAMPLE 5 Standardized Test Practice SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division.

  15. The correct answer is A. ANSWER EXAMPLE 5 Standardized Test Practice Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x3– 2x2– 23x + 60 = (x – 3)(x2 + x – 20) = (x – 3)(x + 5)(x – 4) The zeros are3, –5, and4.

  16. Class/Homework AssignmentWS 5.5 (1-24 mult. of 3)

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