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Test the following polynomials for multiple factors in \(\mathbb{Q}[x]\). (i) \(x^{3}-3 x^{2}+4\) (ii) \(x^{3}-2 x^{2}-x+2\).

Short Answer

Expert verified
(i) Factors: \((x+1)(x-2)^2\). (ii) Factors: \((x-1)(x-2)(x+1)\).

Step by step solution

01

Determine Possible Rational Roots Using Rational Root Theorem

For a polynomial \( p(x) = a_nx^n + \, ... \, + a_0 \), the Rational Root Theorem says that any rational root \( \frac{p}{q} \) must have \( p \) as a factor of \( a_0 \) and \( q \) as a factor of \( a_n \). For \( (i) : x^3 - 3x^2 + 4 \), the possible rational roots are \( \pm 1, \pm 2, \pm 4 \) since the constant term 4 and leading coefficient is 1.
02

Test Possible Rational Roots for (i)

Substitute each possible rational root into the polynomial \( x^3 - 3x^2 + 4 \):\( x = 1: \, 1^3 - 3(1)^2 + 4 = 2 \) (not 0, not a root);\( x = -1: \, (-1)^3 - 3(-1)^2 + 4 = 0 \) (root);Check others, but if one root is found, we can attempt factorization.
03

Factor the Polynomial for (i)

Since \( x = -1 \) is a root, we can perform synthetic or long division of \( x^3 - 3x^2 + 4 \) by \( x + 1 \). The quotient is \( x^2 - 4x + 4 \). Factor this quadratic: \( x^2 - 4x + 4 = (x-2)^2 \). Thus, the polynomial factors are \( (x+1)(x-2)(x-2) \).
04

Determine Possible Rational Roots for (ii)

For \( (ii) : x^3 - 2x^2 - x + 2 \), the possible rational roots are \( \pm 1, \pm 2 \). The constant term is 2, and the leading coefficient 1.
05

Test Possible Rational Roots for (ii)

Substitute each possible rational root into the polynomial \( x^3 - 2x^2 - x + 2 \):\( x = 1: \, 1^3 - 2(1)^2 - 1 + 2 = 0 \) (root);Check others if needed, but if root is found, try factorization.
06

Factor the Polynomial for (ii)

Since \( x = 1 \) is a root, perform synthetic or long division of \( x^3 - 2x^2 - x + 2 \) by \( x-1 \). The quotient is \( x^2 - x - 2 \). Factor the quadratic: \( x^2 - x - 2 = (x-2)(x+1) \). Thus, the polynomial factors are \( (x-1)(x-2)(x+1) \).

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Key Concepts

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

Rational Root Theorem
The Rational Root Theorem provides a helpful method for identifying potential rational solutions of a polynomial equation. Here's how it works:
  • Consider a polynomial equation in the form: \( f(x) = a_nx^n + \, ... \, + a_0 \).
  • The theorem states that any possible rational root, represented as \( \frac{p}{q} \), must have \( p \) as a factor of the constant term \( a_0 \) and \( q \) as a factor of the leading coefficient \( a_n \).
This theorem is especially useful when you need to list out potential roots before testing them individually.
When working with polynomial \( x^3 - 3x^2 + 4 \), the constant term is 4 and the leading coefficient is 1, resulting in potential rational roots of \( \pm 1, \pm 2, \pm 4 \).
For the polynomial \( x^3 - 2x^2 - x + 2 \), the constant term is 2, leading to possible rational roots of \( \pm 1, \pm 2 \). By substituting each value into the polynomial and evaluating the expression, you can determine which values are actual roots.
Synthetic Division
Synthetic division offers a streamlined method for dividing polynomials, particularly when dividing by linear expressions of the form \( x - c \).
Here's a simple synthetic division process:
  • Align the coefficients of the polynomial in descending order of power.
  • Bring down the leading coefficient unchanged.
  • Multiply this coefficient by \( c \) (where \( x-c \) is your divisor) and add it to the next coefficient.
  • Repeat the multiply-and-add process for each coefficient.
  • The final row of coefficients represents the quotient and possible remainder.
In the example of \( x^3 - 3x^2 + 4 \), synthetic division using \( x = -1 \) results in a quotient of \( x^2 - 4x + 4 \).
Applying synthetic division to \( x^3 - 2x^2 - x + 2 \) using \( x = 1 \) gives a quotient of \( x^2 - x - 2 \). This process helps simplify the factorization of polynomials by confirming if a potential root is valid.
Long Division
Long division is another tool for dividing polynomials, similar to numeric division, and is particularly valuable when synthetic division isn't applicable or convenient.
Here’s a step-by-step guide:
  • Divide the first term of the dividend by the first term of the divisor to find the first term of the quotient.
  • Multiply the entire divisor by this term and subtract the result from the original dividend.
  • Bring down the next term from the dividend and repeat the process until all terms are accounted for.
In our polynomial example \( x^3 - 3x^2 + 4 \), long division with \( x+1 \) checks if \( x=-1 \) is indeed a valid root.
For \( x^3 - 2x^2 - x + 2 \), by subtracting \( x-1 \) repeatedly, it helps confirm \( x = 1 \) is a root.
Long division provides a consistent method for breaking down more complex polynomials, confirming divisions, and enabling further factorization.

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Most popular questions from this chapter

Let \(F\) be a field of positive characteristic \(p\) and \(a \in F\) such that \(a\) has no \(p\) th root in \(F\). Prove that \(x^{p}-a \in F[x]\) is irreducible.

Let \(q\) be a prime power, \(t \in \mathbb{N}\) a prime divisor of \(q-1\), and \(a \in \mathbb{F}_{q}^{\times}\). (i) Show that the polynomial \(x^{t}-a \in \mathbb{F}_{q}[x]\) splits into linear factors if \(a\) is a \(t\) th power (ii) Show that \(x^{t}-a\) is irreducible if \(a\) is not a \(t\) th power Hint: Use (i) for the splitting field of \(x^{t}-a\) and consider the constant coefficient of a hypothetical factor \(f \in \mathbb{F}_{q}[x]\) of \(x^{t}-a\). (iii) Derive a formula for the probability that a random binomial \(x^{t}-a\) (that is, for random \(a \in \mathbb{F}_{q}^{\times}\)) is irreducible, and compare it to the probability that a random polynomial of degree \(t\) in \(\mathbb{F}_{q}[x]\) is irreducible.

Prove or disprove: (i) The polynomial \(x^{1000}+2 \in \mathbb{F}_{5}[x]\) is squarefree. (ii) Let \(F\) be a field and \(f, g \in F[x]\). Then the squarefree part of \(f g\) is the product of the squarefree parts of \(f\) and of \(g\).

Compute the squarefree decomposition of the following polynomials in \(\mathbb{Q}[x]\) and in \(\mathbb{F}_{3}[x]\). (i) \(x^{6}-x^{5}-4 x^{4}+2 x^{3}+5 x^{2}-x-2\), (ii) \(x^{6}-3 x^{5}+6 x^{3}-3 x^{2}-3 x+2\), (iii) \(x^{5}-2 x^{4}-2 x^{3}+4 x^{2}+x-2\), (iv) \(x^{6}-2 x^{5}-4 x^{4}+6 x^{3}+7 x^{2}-4 x-4\), (v) \(x^{6}-6 x^{5}+12 x^{4}-6 x^{3}-9 x^{2}+12 x-4\).

Let \(\mathbb{F}_{q}\) be a finite field with \(q\) elements and \(a, b \in \mathbb{F}_{q}^{\times}\)two nonsquares. Prove that \(a b\) is a square.

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