2. (Review) Solving a Polynomial Equation by Factoring 1.Set = 0. 2.Factor. 3.Set each factor equal to zero. (Apply the zero  product principle.) 4. Solve the resulting equations. 5.Check the solutions in the original equation. How do you recognize a “polynomial equation”? 1.6 Other Types of Equations See p 162, check out the graphs and the corresponding functions.

3. Text Example Solve by factoring: 3x 4 = 27x 2. Step 1 Move all terms to one side and obtain zero on the other side. Subtract 27x 2 from both sides 3x 4  x 2  27x 2  27x 2 3x 4  27x 2  Step 2 Factor. 3x 4  27x 2  3x 2 (x 2 - 9)  0 Don’t look at notes, no need to write.

4. Solution cont. Solve by factoring: 3x 4 = 27x 2. Steps 3 and 4 Set each factor equal to zero and solve each resulting equation. 3x 2 = 0orx 2 - 9 = 0 x 2 = 0x 2 = 9 x =  0x =  9 x = 0x =  3 Steps 5 check your solution If time: p 160 #6

5. (Rvw.)Solving RADICAL Equations “IRS Audit” ISOLATE the radical RAISE each side to the same power (match the index) SOLVE the resulting equation “AUDIT” (check), especially if raised to an EVEN power. (ex: \/x = -2)

6. Ex: Solve: Isolate Raise Solve “Audit” Why do we need to know this? Ex: p162#111

7. Ex: Solve: Isolate Raise Solve “Audit”

8. Solving Radical Equations of the Form x m/n = k Assume that m and n are positive integers, m/n is in lowest terms, and k is a real number. 1.ISOLATE the expression with the rational exponent. 2.RAISE both sides of the equation to the n/m power. 3.SOLVE the resulting equation.

9. Solving Radical Equations of the Form x m/n = k cont. If m is even:If m is odd: x m/n = k x m/n = k (x m/n ) n/m = ± k n/m (x m/n ) n/m = k n/m x = ±k n/m x = k n/m It is incorrect to insert the ± when the numerator of the original exponent is odd. An odd index has only one root. 4. “AUDIT” all proposed solutions in the original equation to find out if they are actual solutions or extraneous solutions.

10. Text Example (No need to write, don’t look at notes.) Solve: x 2/3 - 3/4 = -1/2. Isolate x 2/3 by adding 3/4 to both sides of the equation: x 2/3 = 1/4. Raise both sides to the 3/2 power: (x 2/3 ) 3/2 = ±(1/4) 3/2. (The square root of ¼ is ½, cubed is 1/8) (Note: WE take the square (even) root, so we use +/-) x = ±1/8. (Do p 160 # 34: )

11. Some equations that are not quadratic can be written as quadratic equations using an appropriate substitution. Here are some examples. 5t 2 + 11t + 2 = 0t = ________5x 2/3 + 11x 1/3 + 2 = 0 or 5(x 1/3 ) 2 + 11x 1/3 + 2 = 0 t 2 – 8t – 9 = 0t = _______x 4 – 8x 2 – 9 = 0 or (x 2 ) 2 – 8x 2 – 9 = 0 New EquationSubstitutionGiven Equation Equations That Are Quadratic in Form Hint: put the trinomial in descending order, use the variable part of the 2 nd term as t. Square it to see if it works for the variable part of the first term. *** Don’t forget to SUBSTITUTE BACK! *** Do p 160 # 46

12. (Rvw.)Rewriting an Absolute Value Equation without Absolute Value Bars If c is a positive real number and X represents any algebraic expression, then |X| = c is equivalent to X = c or X = -c. (Using numbers: |2| = 2, and |-2| = 2, so if |x| = 2, that means that x = 2 or x = -2) * IMPORTANT* To apply this, we must ISOLATE the absolute value part before setting up cases and dropping the bars.

13. Example (Don’t look at notes, no need to write.) Solve: Answer: 3x-1 = 4 and 3x-1 = -4 3x = 53x = -3 x = 5/3 x = -1 Do on board: Solve | 1 – x | -7 = - 4 and | x + 2|+7 = 3 (Solve: )