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Solving Polynomial Equations

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Presentation on theme: "Solving Polynomial Equations"β€” Presentation transcript:

1 Solving Polynomial Equations
Type Rules Difference of two squares π‘Ž 2 βˆ’ 𝑏 2 =(π‘Ž+𝑏)(π‘Žβˆ’π‘) Perfect Square Trinomials π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 = π‘Ž+𝑏 2 π‘Ž 2 βˆ’2π‘Žπ‘+ 𝑏 2 = π‘Žβˆ’π‘ 2 General trinomials π‘Žπ‘ π‘₯ 2 + π‘Žπ‘‘+𝑏𝑐 π‘₯+𝑏𝑑=(π‘Žπ‘₯+𝑏)(𝑐π‘₯+𝑑) Grouping ax+bx+ay+by = x(a+b)+y(a+b) = (x+y)(a+b) Greatest Common Factor 4 π‘Ž 3 π‘₯ 2 βˆ’8π‘Žπ‘₯=4π‘Žπ‘₯( π‘Ž 2 π‘₯βˆ’2) Sum of Two Cubes π‘Ž 3 + 𝑏 3 =(π‘Ž+𝑏)( π‘Ž 2 βˆ’π‘Žπ‘+ 𝑏 2 ) Difference of Two Cubes π‘Ž 3 βˆ’ 𝑏 3 =(π‘Žβˆ’π‘)( π‘Ž 2 +π‘Žπ‘+ 𝑏 2 )

2 Factoring If a polynomial is not factorable then it is PRIME.

3 Factor Completely 1) 8 𝑐 3 βˆ’125 𝑑 3 2) π‘Ž 6 π‘₯ 2 βˆ’ 𝑏 6 π‘₯ 2
1) 8 𝑐 3 βˆ’125 𝑑 3 2) π‘Ž 6 π‘₯ 2 βˆ’ 𝑏 6 π‘₯ 2 1) (2π‘βˆ’5𝑑)(4 𝑐 2 +10𝑐𝑑+25 𝑑 2 ) 2) π‘₯ 2 (π‘Žβˆ’π‘)( π‘Ž 2 +π‘Žπ‘+ 𝑏 2 )(π‘Ž+𝑏)( π‘Ž 2 βˆ’π‘Žπ‘+ 𝑏 2 )

4 Quadratic form Equations with degrees higher than 2 (or lower than 2) can sometimes be β€œreduced” or converted to quadratic equations π‘Ž π‘₯ 2 +𝑏π‘₯+𝑐=𝑦. π‘₯ 4 βˆ’4 π‘₯ 2 +3=0 we can substitute w = π‘₯ 2 and rewrite the equation as: 𝑀 2 βˆ’4𝑀+3=0 and then solve the new equation (w-3)(w-1)=0 so w=3, w=1 but w= π‘₯ 2 π‘ π‘œ π‘₯ 2 =3 π‘Žπ‘›π‘‘ π‘₯ 2 =1 so x=Β± 3 ,Β±1 Or π‘₯ βˆ’4 (π‘₯ 2 )+3=0 (π‘₯ 2 -3) (π‘₯ 2 -1)=0

5 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 – make sure the answer makes sense.

6 substitution 2 π‘₯ 4 βˆ’5 π‘₯ 2 βˆ’12=0 Substitute y= π‘₯ 2 2 𝑦 2 βˆ’5π‘¦βˆ’12=0 (2y+3)(y-4) Substitute back (2 π‘₯ 2 +3)( π‘₯ 2 -4)=0 2 π‘₯ 2 +3=0 π‘₯ π‘₯ 2 =βˆ’3 π‘₯ 2 =4 No real solution +2,-2

7 Practice What would we substitute in the following equations? 1. 2.
Now solve problem 1. Remember to check the solutions!

8 Solution 1) x = {Β±1,Β±2}

9 More challenging!! What would we substitute: 1Β± 2 , 1 2 Β±

10 Worksheets Factoring and Word Problems


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