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Section 4.3 Dividing Polynomials

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1 Section 4.3 Dividing Polynomials
Honors Algebra 2 Section 4.3 Dividing Polynomials

2 Warm Up How do you do the following problem without a calculator?
3,896Γ·5 How do you divide the following? (12 π‘₯ 3 βˆ’6 π‘₯ 2 +8π‘₯)Γ·2π‘₯

3 Long Division (Retro Math)

4 You can divide polynomials
using the same process!!!! Let’s look at ( π‘₯ 3 βˆ’ π‘₯ 2 βˆ’2π‘₯+8)Γ·(π‘₯βˆ’1) Try this! ( π‘₯ 4 +2 π‘₯ 2 βˆ’π‘₯+5)Γ·( π‘₯ 2 βˆ’π‘₯+1) Put in any missing terms- we don’t have an π‘₯ 3 term so put in 0 π‘₯ 3 .

5 Synthetic division- shortcut for dividing polynomials by binomials of the form π‘₯βˆ’π‘˜

6 Let’s look at ( π‘₯ 3 βˆ’ π‘₯ 2 βˆ’2π‘₯+8)Γ·(π‘₯βˆ’1) again! π‘˜=1 and the polynomial is divided by a binomial. YEAH! Shortcut works! 1Λ© βˆ’ βˆ’ ↓ βˆ’2 remainder

7 The pattern for synthetic division is
Bring down Multiply Add Add until you get to the end You use these bottom row numbers as your coefficients(start the variable exponent down one number if the binomial had x and two numbers if the binomial had π‘₯ 2 ) and constant of your polynomial quotient and the last number is the remainder.

8 Use synthetic division with the following problems.
#1 (3 π‘₯ 3 βˆ’7 π‘₯ 2 +6π‘₯+8)Γ·(π‘₯βˆ’1) #2 (5 π‘₯ 5 βˆ’3 π‘₯ 3 +8)Γ·(π‘₯+2)

9 The Remainder Theorem If the polynomial is 𝑓(π‘₯) is divided by π‘₯βˆ’π‘˜, then the remainder is π‘Ÿ=𝑓(π‘˜) Use synthetic division to find this value.

10 Use synthetic division to find the following function values.
#1 𝑓 βˆ’2 π‘“π‘œπ‘Ÿ 𝑓 π‘₯ =3 π‘₯ 3 βˆ’4 π‘₯ 2 +5π‘₯βˆ’11 #2 𝑓 6 π‘“π‘œπ‘Ÿ 𝑓 π‘₯ =5 π‘₯ 4 βˆ’2 π‘₯ 2 +12

11 Assignment #17 Pg. 177 #4, 5-23 odd, 25, 27, 35


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