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5.5, Day 2 More Synthetic Division

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1 5.5, Day 2 More Synthetic Division

2 If 𝑓 π‘₯ = π‘₯ 3 + π‘₯ 2 βˆ’16π‘₯βˆ’16 has a zero of 4, what would the other zeros be?
If 4 is a zero, that means that x – 4 is a factor and I should be able to synthetically divide by 4 and find the other factors and from there find the other zeros. Good, we expected a remainder of zero, and now we need to factor π‘₯ 2 +5π‘₯+4 . 4 20 16 1 5 4 π‘₯ 2 +5π‘₯+4=(π‘₯+1)(π‘₯+4), so the other zeros are -1 and -4 What would f(-4) be for the above equation? zero

3 If f(-1) = 0, completely factor 𝑓 π‘₯ = π‘₯ 3 +2 π‘₯ 2 βˆ’5π‘₯βˆ’6.
If f(-1) = 0, then x + 1 is a factor, and I can synthetically divide by -1 and finish factoring. -1 -1 6 1 1 -6 π‘₯ 2 +π‘₯βˆ’6=(π‘₯+3)(π‘₯βˆ’2) Therefore the complete factorization of π‘₯ 3 + 2π‘₯ 2 βˆ’5π‘₯βˆ’6 is π‘₯+1 π‘₯+3 π‘₯βˆ’2 .

4 If f(-2) = 0, completely factor 𝑓 π‘₯ = 7π‘₯ 3 βˆ’10 π‘₯ 2 βˆ’39π‘₯+18.
-14 48 -18 7 -24 9 7 π‘₯ 2 βˆ’24π‘₯+9=(7π‘₯βˆ’3)(π‘₯βˆ’3) The complete factorization of 7π‘₯ 3 βˆ’ 10π‘₯ 2 βˆ’39π‘₯ is π‘₯+2 7π‘₯βˆ’3 π‘₯βˆ’3 . What are all the zeros of 7π‘₯ 3 βˆ’10 π‘₯ 2 βˆ’39π‘₯+18 ? Zero’s are -2, , and 3 Factor theorem---A polynomial f(x) has a factor of (x – k), if and only if f(k) = 0 .

5 A company’s profit C (in thousands of dollars) can be modeled by 𝐢=βˆ’ 5π‘₯ 3 +6 π‘₯ 2 +15π‘₯, where x is the number of items produced in thousands. The profit is $14,000 for producing 2000 items. What other number of items would produce about the same profit? When C = 14, there is a zero of 2. 0= βˆ’5π‘₯ 3 + 6π‘₯ 2 +15π‘₯βˆ’14 -10 -8 14 -5 -4 7 The expression that is left, βˆ’5 π‘₯ 2 βˆ’4π‘₯+7, does not factor, so use the quadratic formula to find an approximate zero that seems appropriate. Approx , or 850 items!

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