Thus p x x a q x thus x a is a factor of

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Unformatted text preview: ( x 5 − 8 x 4 + 12 x 3 + 23 x 2 − 16 x − 37 ) ÷ ( x − 3 ) , we have that 5 4 3 2 8 + Coefficients of x − x 12 x + 23 x − 16x − 37 1 −8 12 23 − 16 − 37 3 1 − 15 −9 42 78 −5 −3 14 26 3 41 Thus, the remainder is 41. Thus, h ( 3 ) = 41 . Theorem (The Factor Theorem) Let p be a polynomial. The expression x − a is a factor of p( x ) if and only if p( a ) = 0 . Proof ( ⇒ ) Suppose that x − a is a factor of p( x ) . Then the remainder upon division by x − a must be zero. By the Remainder Theorem, p( a ) = 0 . ( ⇐ ) Suppose that p ( a ) = 0 . By the Remainder Theorem, we have that p ( x ) = ( x − a ) q( x ) + p ( a ) . Thus, p ( x ) = ( x − a ) q( x ) . Thus, x − a is a factor of p( x ) . Example Show that x + 4 is a factor of p ( x ) = 5 x 3 + 12 x 2 − 20 x + 48 . We will use the Factor Theorem and show that p( − 4 ) = 0 . We will use the Remainder Theorem and synthetic division to find p( − 4 ) . 3 2 Coeff x− of 5 +12 20 x 48 x + 5 12 − 20 48 − 20 5 32 − 48 −8 12 −4 0 Copyrighted by James D. Anderson, The University of Toledo www.math.utoledo.edu/~anderson/1320 Thus, p( − 4 ) = 0 . Thus, by the Factor Theorem, x + 4 is a factor of p ( x ) = 5 x 3 + 12 x 2 − 20 x + 48 . NOTE: The third row in the synthetic division gives us the coefficients of the other factor starting with x 2 . Thus, the other factor is 5 x 2 − 8 x + 12 . Thus, we have that 5 x 3 + 12 x 2 − 20 x + 48 = ( x + 4 ) ( 5 x 2 − 8 x + 12 ) . Example Show that t − 6 is not a factor of q( t ) = 2 t 4 − 7 t 2 + 15 . We will use the Factor Theorem and show that q( 6 ) ≠ 0...
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This note was uploaded on 02/11/2014 for the course MATH 1320 taught by Professor Staff during the Spring '08 term at Toledo.

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