120224 Advanced factoring methods

Example solve the equation x 4 10 x 2 9 0 solution x

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Example: Solve the equation x 4 – 10 x 2 + 9 = 0. Solution: x 4 – 10 x 2 + 9 = 0 steps as in the previous example leading to ( x +1)( x –1)( x +3)( x –3) = 0 x = –1, x = 1, x = –3, x = 3. You try it 3. Factor x 6 – 7 x 3 – 8. 4. Solve x 4 – 9 x 2 + 18 = 0.
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Name Honors Algebra 2 February 24, 2012 “Advanced factoring methods” page 3 C. Factoring by completing the square This method applies mainly to fourth-degree polynomials that only have x 4 , x 2 , and constant terms. The approach is to add and subtract a multiple of x 2 that creates a perfect square. Then the polynomial can be expressed as a difference of perfect squares, and factored using the a 2 b 2 = ( a + b )( a b ) method. Example: Factor x 4 + 2 x 2 + 9. Solution: The perfect square involving x 4 + 9 would be x 4 + 6 x + 9, so begin by adding and subtracting to create this perfect square. x 4 + 2 x 2 + 9 = x 4 + 2 x 2 + 4 x 2 + 9 – 4 x 2 = x 4 + 6 x 2 + 9 – 4 x 2 = ( x 2 + 3) 2 – (2 x ) 2 = ( ( x 2 + 3) + (2 x ) ) ( ( x 2 + 3) – (2 x ) ) = ( x 2 + 2 x + 3) ( x 2 – 2 x + 3). You try it 5. Factor x 4 + 6 x 2 + 25. 6. Factor x 4 – 5 x 2 + 4.
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Name Honors Algebra 2 February 24, 2012 “Advanced factoring methods” page 4 Mixed practice Directions: Factor these polynomials as completely as possible. Various methods are needed
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