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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online