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Solving Polynomial Equations
Type Rules Difference of two squares π 2 β π 2 =(π+π)(πβπ) Perfect Square Trinomials π 2 +2ππ+ π 2 = π+π 2 π 2 β2ππ+ π 2 = πβπ 2 General trinomials ππ π₯ 2 + ππ+ππ π₯+ππ=(ππ₯+π)(ππ₯+π) Grouping ax+bx+ay+by = x(a+b)+y(a+b) = (x+y)(a+b) Greatest Common Factor 4 π 3 π₯ 2 β8ππ₯=4ππ₯( π 2 π₯β2) 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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Factoring If a polynomial is not factorable then it is PRIME.
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Factor Completely 1) 8 π 3 β125 π 3 2) π 6 π₯ 2 β π 6 π₯ 2
1) 8 π 3 β125 π 3 2) π 6 π₯ 2 β π 6 π₯ 2 1) (2πβ5π)(4 π 2 +10ππ+25 π 2 ) 2) π₯ 2 (πβπ)( π 2 +ππ+ π 2 )(π+π)( π 2 βππ+ π 2 )
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Quadratic form Equations with degrees higher than 2 (or lower than 2) can sometimes be βreducedβ or converted to quadratic equations π π₯ 2 +ππ₯+π=π¦. π₯ 4 β4 π₯ 2 +3=0 we can substitute w = π₯ 2 and rewrite the equation as: π€ 2 β4π€+3=0 and then solve the new equation (w-3)(w-1)=0 so w=3, w=1 but w= π₯ 2 π π π₯ 2 =3 πππ π₯ 2 =1 so x=Β± 3 ,Β±1 Or π₯ β4 (π₯ 2 )+3=0 (π₯ 2 -3) (π₯ 2 -1)=0
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Equations Reducible to Quadratic Form
Changing an equation that is not in quadratic form into quadratic form is called reducing the equation to quadratic form. Steps Change any verbal equation into an algebraic equation. Determine what, if any, substitution can be made to change the equation into a quadratic equation. Solve using any method of solving a quadratic equation Reverse the substitution to get the answer to the original problem. CHECK β make sure the answer makes sense.
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substitution 2 π₯ 4 β5 π₯ 2 β12=0 Substitute y= π₯ 2 2 π¦ 2 β5π¦β12=0 (2y+3)(y-4) Substitute back (2 π₯ 2 +3)( π₯ 2 -4)=0 2 π₯ 2 +3=0 π₯ π₯ 2 =β3 π₯ 2 =4 No real solution +2,-2
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Practice What would we substitute in the following equations? 1. 2.
Now solve problem 1. Remember to check the solutions!
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Solution 1) x = {Β±1,Β±2}
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More challenging!! What would we substitute: 1Β± 2 , 1 2 Β±
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Worksheets Factoring and Word Problems
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