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Opener Perform the indicated operation.
6β4 π₯ 2 +5 π₯ 3 β 7 π₯ 2 β π₯ 4 β12 π₯ 3 (π₯β6)(2π₯+1)(5π₯β3) Use synthetic substitution to evaluate π π₯ = π₯ 3 β2π₯β1 when π₯=3 Solve 6 π₯ 2 +π₯=15
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Factoring and Solving Polynomial Equations
Notes: 6.4 Factoring and Solving Polynomial Equations
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Factoring Patterns Sum of Two Cubes π 3 + π 3 =(π+π)( π 2 βππ+ π 2 )
Difference of Two Cubes π 3 β π 3 =(πβπ)( π 2 +ππ+ π 2 )
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Problem Set 1: Factoring the Sum or Difference of Two Cubes
π₯ 3 +27 16 π’ 5 β250 π’ 2 125+ π₯ 3 64 π 4 β27π
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Problem Set 2: Factor by Grouping
π₯ 3 β2 π₯ 2 β9π₯+18 π π₯ 2 +3π+3π+π π₯ 2 π₯ 2 π¦ 2 β3 π₯ 2 β4 π¦ 2 +12
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Problem Set 3: Factoring Polynomials in Quadratic Form
81 π₯ 4 β16 4 π₯ 6 β20 π₯ π₯ 2 25 π₯ 4 β36 π 2 π 2 β8π π π 4
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Problem Set 4: Solving a Polynomial Equation
2 π₯ 5 +24π₯=14 π₯ 3 2 π¦ 5 β18π¦=0
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