1 / 10

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

ANSWER. 2. Use synthetic substitution to evaluate f ( x ) = x 3 + x 2 – 3 x – 10 when x = 2. – 4. ANSWER. VOCAB. Old school long division terminology. SOLUTION. EXAMPLE 1. Use polynomial long division. Divide f ( x ) = 3 x 4 – 5 x 3 + 4 x – 6 by x 2 – 3 x + 5. SOLUTION.

dunnjamie
Télécharger la présentation

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

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. VOCAB Old school long division terminology SOLUTION

  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. quotient ) 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 Multiply divisor by 3x4/x2 = 3x2 Subtract. Bring down next term. 4x3 – 15x2 + 4x Multiply divisor by4x3/x2 = 4x Subtract. Bring down next term. – 3x2 – 16x – 6 Multiply divisor by– 3x2/x2 = – 3 – 25x + 9

  5. 3x4 – 5x3 + 4x – 6 – 25x + 9 ANSWER = 3x2 + 4x – 3 + x2 – 3x + 5 x2 – 3x + 5 = 3x4 – 5x3 + 4x – 6 EXAMPLE 1 Use polynomial long division CHECK You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. The result should be the dividend. (3x2 + 4x – 3)(x2 – 3x + 5) + (– 25x + 9) = 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9 = 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9

  6. 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. x2 + 7x + 7 Multiply divisor byx3/x = x2. 7x2 – 7x Subtract. Multiply divisor by7x2/x = 7x. 7x + 2 Subtract. Multiply divisor by7x/x = 7. 16

  7. 2x4 + x3 + x – 1 x3 – x2 +4x – 10 – 30 – 18x + 7 ANSWER ANSWER = (2x2 – 3x + 8) + = (x2 – 3x +10) + x2 + 2x – 1 x + 2 x2 + 2x – 1 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)

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

  9. – 2 3 – 4 – 28 – 16 – 6 20 16 3 – 10 – 8 0 EXAMPLE 4 Factor a polynomial Factorf (x) = 3x3– 4x2– 28x – 16completely given that x + 2is a factor. SOLUTION Because x + 2 is a factor of f (x), you know that f (–2)= 0. Use synthetic division to find the other factors.

  10. 4x3 + x2– 3x + 7 x3 + 4 x2–x – 1 11 9 = 4x2+ 5x + 2 + = x2+ x – 4 + ANSWER ANSWER x + 3 x + 3 x – 1 x – 1 for Examples 3 and 4 GUIDED PRACTICE Divide using synthetic division. 3. (x3 + 4x2–x – 1) ÷ (x + 3) 4. (4x3 + x2– 3x + 7) ÷ (x – 1)

More Related