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SolvingPolynomial Equations๏ Solve by Factoring๏ Identify solutions on Graph๏ Solve by Graphing
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Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).
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Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic). ★ There are 3 ways to solve Polynomial Equations
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Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic). ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property
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Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic). ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property (2) Using the graphing calculator to graph
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Solving Polynomial Equations★ Solving a polynomial equation is the same as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic). ★ There are 3 ways to solve Polynomial Equations (1) Using factoring and the zero product property (2) Using the graphing calculator to graph (3) Using Synthetic Division (separate notes)
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Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial.★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.★ Some solutions may be irrational or even imaginary.
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Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.★ Some solutions may be irrational or even imaginary.
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Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.
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Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.★ It is possible for a polynomial equation to have fewer solutions (or none at all).
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Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.
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Total Number of Solutions★ The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial. ( ) 3 f x = x + x − x +12 Degree is 3. Therefore 3 or fewer solutions possible.★ It is possible for a polynomial equation to have fewer solutions (or none at all).★ The degree of the polynomial gives you the maximum number of solutions that are theoretically possible.★ Some solutions may be irrational or even imaginary.
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier.★ Factor completely!★ Set each factor equal to 0 and solve.
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely!★ Set each factor equal to 0 and solve.
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely!
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here.
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0★ Set each factor equal to 0 and solve.
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0★ Set each factor equal to 0 and solve. 3x = 0 ( x − 2) = 0 ( x + 2) = 0
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Solving by Factoring★ When using factoring, remember to always look for a GCF (Greatest Common Factor) ﬁrst. This will help make the remaining factoring easier. 3x 3 − 12x = 0 GCF ( ) 3x x 2 − 4 = 0★ Factor completely! Keep going here. 3x ( x − 2 ) ( x + 2 ) = 0★ Set each factor equal to 0 and solve. 3x = 0 ( x − 2) = 0 ( x + 2) = 0 x=0 x=2 x = −2
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up to 4 solutions.
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 )
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2x −9=0
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 x =9
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 x =9 x=± 9
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 x =9 x=± 9 x = ±3
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ±3
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) 2 2x −9=0 x +3= 0 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3 x = ±i 3
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Example: Solve by Factoring 4 2 x − 6x = 27 Degree is 4 so can up 4 2 to 4 solutions. x − 6x − 27 = 0 (x 2 )( −9 x +3 =0 2 ) Solutions: 2x −9=0 2 x +3= 0 {±3, ±i 3} 2 2 x =9 x = −3 x=± 9 x = ± −3 x = ±3 x = ±i 3
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions.
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 x = −3
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 2 x =1 x = −3
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 ( ) x − 1 ( x + 3) = 0 2 2 x −1= 0 x+3= 0 2 x =1 x = −3 x = ±1
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Try this: Solve by Factoring 3 2 x + 3x − x − 3 = 0 Degree is 3 so can up to 3 solutions. ( ) x + 3x + ( −x − 3) = 0 3 2 x ( x + 3) − 1( x + 3) = 0 2 Solutions: ( ) x − 1 ( x + 3) = 0 2 {1, −1, −3} 2 x −1= 0 x+3= 0 2 x =1 x = −3 x = ±1
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Try this: Solve by Factoring 3 2 x + x − 4x = 0
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Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions.
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Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions. ( 2 x x +x−4 =0 )
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Try this: Solve by Factoring 3 2 x + x − 4x = 0 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 )x=0
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Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2x=0 x +x−4=0
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Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula
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Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1)
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Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1) −1 ± 17 x= 2
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Try this: Solve by Factoring 3 x + x − 4x = 0 2 Degree is 3 so can up to 3 solutions. ( x x +x−4 =02 ) 2 Not factorable so usex=0 x +x−4=0 the quadratic formula a = 1; b = 1; c = −4 −1 ± 12 − 4 (1) ( −4 ) x= 2 (1) Solutions: −1 ± 17 −1 ± 17 x= 0, 2 2
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Try this: Solve by Factoring f ( x ) = x + 64 3
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions.
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + (4) 3
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + (4) 3 Sum of cubes. Apply the formula.
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 30 = ( x + 4 ) ( x − 4x + 16 ) 2
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 30 = ( x + 4 ) ( x − 4x + 16 ) 2x+4=0
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 30 = ( x + 4 ) ( x − 4x + 16 ) 2 2x+4=0 x − 4x + 16 = 0
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 2 x+4=0 x − 4x + 16 = 0−4 = x
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square.−4 = x
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 x − 2 = ± −12
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 x − 2 = ± −12 x = 2 ± 2i 3
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Try this: Solve by Factoring f ( x ) = x + 64 3 Degree is 3 so can up to 3 solutions. 3 0 = x + ( 4 ) Sum of cubes. Apply the formula. 3 0 = ( x + 4 ) ( x − 4x + 16 ) 2 Not factorable so 2 x+4=0 x − 4x + 16 = 0 use completing the square. 2−4 = x x − 4x + ( −2 ) = −16 + 4 2 2 ( x − 2 ) = −12 Solutions: x − 2 = ± −12 {−4, 2 ± 2i 3} x = 2 ± 2i 3
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Practice Time!★ Follow this link to practice solving polynomial equations using Factoring.
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Solutions by Observing the Graph★ The degree of the function tells you the maximum number of solutions possible.
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Solutions by Observing the Graph★ The degree of the function tells you the maximum number of solutions possible.★ The real solutions are where the function crosses or touches the x-axis.
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Solutions by Observing the Graph★ The degree of the function tells you the maximum number of solutions possible.★ The real solutions are where the function crosses or touches the x-axis.★ The graph below has 4 solutions because it crosses the x-axis in 4 places. Notice 2 are positive real numbers and 2 are negative real numbers.
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You try: Find all real zeros on the graph.★ The real zeros for the graph below are {−1, 2, 5}
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Practice Time!★ Follow this link to practice solving polynomial equations using Factoring.
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Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.
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Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.★ Graph the right side of the equation in Y2.
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Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.★ Graph the right side of the equation in Y2.★ Find all the points the two graphs intersect. The x- coordinate is the solution.
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Solving by Graphing in Calculator★ Graph the left side of the equation in Y1.★ Graph the right side of the equation in Y2.★ Find all the points the two graphs intersect. The x- coordinate is the solution.★ If you are given a function such as f(x) = x2 - 1, use zero for f(x). So Y1 = 0 and Y2 = x2 - 1. The ﬁnd all the intersections.
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Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)
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Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
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Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
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Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
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Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1)
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Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1) Solutions: {−2,1}
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Example: Solve by Graphing 2 f ( x ) = 0.25 ( x + 2 ) ( x − 1)Y1 = 0 2Y 2 = 0.25 ( x + 2 ) ( x − 1) Solutions: {−2,1} Notice if you used the zero product property, x = 1 would have occurred twice. We say 1 has multiplicity of 2.
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Example: Solve by Graphing 3 2 4x − 8x = x − 2
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Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
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Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
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Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
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Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
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Example: Solve by Graphing 3 2 4x − 8x = x − 2 3 2Y1 = 4x − 8xY2 = x − 2
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