/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 $$\text { Is } x+b \text { a fac... [FREE SOLUTION] | 91Ó°ÊÓ

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$$\text { Is } x+b \text { a factor of } x^{3}+(2 b-a) x^{2}+\left(b^{2}-2 a b\right) x-a b^{2} ?$$

Short Answer

Expert verified
Yes, \( x + b \) is a factor of the polynomial. By substituting \( x = -b \), we verify the result is zero, confirming \( x + b \) as a factor.

Step by step solution

01

Understand the Factor Theorem

The factor theorem states that \( x + b \) is a factor of a polynomial \( P(x) \) if and only if \( P(-b) = 0 \). We will use this theorem to check if \( x + b \) is a factor of the given polynomial.
02

Define the Polynomial

The given polynomial is \( P(x) = x^3 + (2b-a)x^2 + (b^2 - 2ab)x - ab^2 \). We need to evaluate this polynomial at \( x = -b \).
03

Substitute and Simplify

Substitute \( x = -b \) into the polynomial: \[P(-b) = (-b)^3 + (2b-a)(-b)^2 + (b^2 - 2ab)(-b) - ab^2 \] Simplify each term: - \((-b)^3 = -b^3\)- \((2b-a)(-b)^2 = (2b-a)b^2 = 2b^3 - ab^2\)- \((b^2-2ab)(-b) = -b^3 + 2ab^2\)- \(-ab^2\) is unchanged.
04

Combine and Verify

Combine the simplified terms: \[P(-b) = -b^3 + 2b^3 - ab^2 - b^3 + 2ab^2 - ab^2\]Simplify: \[P(-b) = (-b^3 - b^3 + 2b^3) + (-ab^2 + 2ab^2 - ab^2) = 0\] since both groups add to zero.
05

Conclude the Solution

Since \( P(-b) = 0 \), by the factor theorem, \( x + b \) is indeed a factor of the polynomial \( x^3 + (2b-a)x^2 + (b^2 - 2ab)x - ab^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Factorization
Polynomial factorization is a crucial skill in mathematics that involves breaking down a polynomial into smaller, simpler components, called factors, which when multiplied together, yield the original polynomial. It is similar to factoring numbers but involves variables and coefficients. In our example, we are tasked to check if \( x + b \) is a factor of a given polynomial using the Factor Theorem.The Factor Theorem is a fundamental theorem in algebra that provides a useful method to verify whether a linear expression, like \( x + b \), is a factor of a polynomial. According to the theorem, a polynomial \( P(x) \) is divisible by \( x - c \) if and only if \( P(c) = 0 \). Thus, finding the factors involves substituting potential roots into the polynomial and checking if they satisfy the equation by resulting in zero. Discovering the factors of a polynomial not only simplifies solving equations but also helps analyze the polynomial's roots and behavior.
Polynomial Evaluation
Polynomial evaluation involves calculating the value of a polynomial for a given value of \( x \). This process is pivotal when applying the Factor Theorem. In our exercise, we evaluated the polynomial \( P(x) = x^3 + (2b-a)x^2 + (b^2 - 2ab)x - ab^2 \) by substituting \( x = -b \). This substitution allows us to determine whether \( x + b \) is a factor or not. Here are the key steps in evaluating the polynomial:
  • Substitute \( x = -b \) into each term of the polynomial.
  • Simplify each term by combining like terms and applying algebraic rules.
  • Combine all terms and check if the final sum equals zero.
By ensuring \( P(-b) = 0 \), as in this exercise, the Factor Theorem confirms that \( x + b \) is indeed a factor. This skill of polynomial evaluation is not only important in factorization but also in various applications across mathematics and sciences.
Precalculus Problem Solving
Precalculus problem solving often involves understanding and applying complex mathematical concepts, such as polynomial factorization and evaluation, to tackle challenging exercises. This branch of mathematics equips students with essential skills that act as a stepping stone towards calculus. In the given exercise, the key problem-solving strategy was to apply the Factor Theorem, combined with polynomial evaluation, to assess factorability. Here’s how it unfolds in precalculus:
  • Recognize the form of the polynomial and the expression you are testing as a factor.
  • Apply the Factor Theorem to set up your evaluation.
  • Thoroughly calculate through substitution, simplifying each step to avoid errors.
  • Analyze the results to draw conclusions about the polynomial’s factors.
By refining these problem-solving skills, students enhance their ability to dissect and solve complex polynomial equations efficiently. Mastery of these techniques offers a strong foundation for future mathematical studies, especially those involving calculus.

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Most popular questions from this chapter

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