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- 1. Part 2: Synthetic Division & The Remainder Theorem
- 2. Synthetic Division Synthetic Division is a process that simplifies long division, but it can only be used when dividing a polynomial by a linear factor of the form x – a.
- 3. Synthetic Division 1. Write the polynomial in standard form, including zero coefficients where appropriate 2. Set up: use the opposite sign of a (this allows us to add throughout the process) and write the coefficients of the polynomial. 3. Bring down the first coefficient 4. Multiply the coefficient by the divisor. Add to the next coefficient. 5. Continue multiplying and adding through the last coefficient. 6. Write the quotient and remainder. The remainder will be the last sum.
- 4. Example: Divide using synthetic division (x 3 − 57 x + 56 ) ÷ ( x − 7 )
- 5. Example: Divide using synthetic division (x 3 − 14 x + 51x − 54 ) ÷ ( x + 2 ) 2
- 6. Example: Using synthetic division to solve a problem The polynomial x 3 + 7 x 2 − 38 x − 240 expresses the volume, in cubic inches, of the shadow box shown. 1. What are the dimensions of the box? Hint: the length is greater than the height (or depth) 2. If the width of the box is 15 in, what are the other dimension?
- 7. The Remainder Theorem The Remainder Theorem provides a quick way to find the remainder of a polynomial long-division problem. If you divide a polynomial P(x) of degreen ≥ 1 by x − a then the remainder is P(a)
- 8. Example: Evaluating a Polynomial Given that P ( x ) = x − 2x − x + 2 5 3 2 what is P(3)? By the remainder theorem, P(3) is the remainder when you divide P(x) by x – 3.
- 9. Example: Evaluating a Polynomial Given that P ( x ) = x − 3 x − 28 x + 5 x + 20 5 4 3 what is P(─ 4)? By the remainder theorem, P(─ 4) is the remainder when you divide P(x) by x + 4.
- 10. Homework P308 #21 – 39 odd, 40 – 43 all, 53 – 56 all, 57 – 61 odd