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Chapter 14: Problem 25

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鈥檚 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, and let \(f, g_{1}, \ldots, g_{m} \in F[x]\) be monic nonconstant polynomials. Recall that \(\left(g_{1}, \ldots, g_{m}\right)\) is the squarefree decomposition of \(f\) if \(f=g_{1} g_{2}^{2} \cdots g_{m}^{m}\), each \(g_{i}\) is squarefree, the \(g_{i}\) are pairwise coprime, and \(g_{m} \neq 1\). (i) Prove that there is a unique decomposition \(f=h_{1} \cdots h_{m}\) such that each \(h_{i}\) is monic, nonconstant, and squarefree, \(h_{i} \mid h_{i-1}\) for \(2 \leq i \leq m\), and \(h_{m} \neq 1\). (ii) Give both decompositions for \(f=x^{4}(x+1)^{3}\). (iii) Express the \(h_{i}\) in terms of the \(g_{i}\) and vice versa, and show that both conversions can be computed in time \(O(\mathrm{M}(n))\) if \(n=\operatorname{deg} f\).

Show that \(-1\) is a square in \(\mathbb{F}_{q}^{\times}\)for an odd prime power \(q\) if and only if \(q \equiv 1 \bmod 4\)

If \(G\) is a group and \(a, b \in G\), then \(b\) is a square root of \(a\) if \(b^{2}=a\). (i) Prove that every element of a cyclic group \(G\) has at most two square roots. (ii) Find a counterexample to (i) when \(G\) is not cyclic.

Let \(f \in \mathbb{Z}[x]\) be of degree \(n\) and max-norm \(\|f\|_{\infty}=A\), and \(f=(u x+v) g\), with nonzero \(u, v \in \mathbb{Z}\) and \(g=\sum_{0 \leq i1\).

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.
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