Use (2x3−3x2−11x+6)/(x−3)=2x2+3x−2 to factor 2x3-3x2-11x+6 completely.

Ch. 3 - Polynomial and Rational Functions

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

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Blitzer 8th Edition

Ch. 3 - Polynomial and Rational Functions

Problem 115

Chapter 4, Problem 115

Use (2x³−3x²−11x+6)/(x−3)=2x²+3x−2 to factor 2x³-3x²-11x+6 completely.

Verified step by step guidance

Recognize that the given expression is a division of polynomials: \( \frac{2x^3 - 3x^2 - 11x + 6}{x - 3} = 2x^2 + 3x - 2 \). This means \(x - 3\) is a factor of the polynomial \(2x^3 - 3x^2 - 11x + 6\).

Rewrite the original polynomial as a product of the divisor and the quotient: \(2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x^2 + 3x - 2)\).

Focus on factoring the quadratic polynomial \(2x^2 + 3x - 2\) completely. To do this, look for two numbers that multiply to \(2 \times (-2) = -4\) and add to \(3\).

Use these two numbers to split the middle term \(3x\) into two terms, then factor by grouping. This will help break down \(2x^2 + 3x - 2\) into the product of two binomials.

Combine the factor \(x - 3\) with the factored form of \(2x^2 + 3x - 2\) to write the complete factorization of \(2x^3 - 3x^2 - 11x + 6\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polynomial Division

Polynomial division is a method used to divide one polynomial by another, similar to long division with numbers. It helps simplify expressions and find factors by expressing a polynomial as a product plus a remainder. In this problem, dividing by (x−3) helps break down the cubic polynomial.

Factoring Polynomials

Factoring involves rewriting a polynomial as a product of simpler polynomials or factors. This process is essential for solving polynomial equations and simplifying expressions. After division, the quotient and divisor can be used to express the original polynomial as a product of factors.

Using the Remainder and Factor Theorems

The Factor Theorem states that if a polynomial f(x) divided by (x−a) leaves a remainder of zero, then (x−a) is a factor of f(x). The Remainder Theorem helps find the remainder quickly. Here, since division by (x−3) yields a polynomial with no remainder, (x−3) is a factor.