Precalc0006to0010-page14

Precalc0006to0010-page14 - ( ) 2 = x 6 . 2 x 6 x 3 1 = 2 u...

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

View Full Document Right Arrow Icon
(Section 0.7: Factoring Polynomials) 0.7.9 PART E: FACTORING EXPRESSIONS IN QUADRATIC FORM An expression is in quadratic form ± It can be expressed as au 2 + bu + c after performing a u substitution , where a ± 0 , and a , b , and c are real coefficients. • The term “quadratic form” is defined differently in higher math; that definition requires each term to have degree 2. Example 6 (Factoring an Expression in Quadratic Form) Factor 2 x 6 ± x 3 ± 1 over the integers. § Solution The trinomial 2 x 6 ± x 3 ± 1 is in quadratic form , because the exponent on x in the first term is twice that in the second term (6 is twice 3), and the third term is a constant. We will use the substitution u = x 3 , the power of x in the “middle” term. Then, u 2 = x 3
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( ) 2 = x 6 . 2 x 6 x 3 1 = 2 u 2 u 1 Now, factor as usual. = 2 u + 1 ( ) u 1 ( ) Substitute back. Replace u with x 3 . = 2 x 3 + 1 ( ) x 3 1 ( ) With practice, the substitution process can be avoided. Either way, we are not done yet! It is true that 2 x 3 + 1 ( ) is prime over the integers; Chapter 2 will help us verify that. However, x 3 1 ( ) is not prime, because we can apply the Difference of Two Cubes formula. 2 x 3 + 1 ( ) x 3 1 ( ) = 2 x 3 + 1 ( ) x 1 ( ) x 2 + x + 1 ( ) This is factored completely over the integers. The Test for Factorability can be used to show that the trinomial factor x 2 + x + 1 ( ) is prime , as expected....
View Full Document

This document was uploaded on 12/29/2011.

Ask a homework question - tutors are online