Chapter 4: Problem 68
Use the rational zeros theorem to completely factor \(P(x)\) into linear factors. (Hint: Not all zeros of \(P(x)\) are rational. $$P(x)=5 x^{4}+8 x^{3}-19 x^{2}-24 x+12$$
Short Answer
Expert verified
\(P(x) = (x-1)(x-2)(5x^2 + 23x + 15)\)
Step by step solution
01
Identify the possible rational roots
According to the Rational Root Theorem, any rational root of the polynomial \(P(x)\) in the form \(\frac{p}{q}\) has \(p\) as a factor of the constant term (12) and \(q\) as a factor of the leading coefficient (5). So the possible rational roots are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{3}{5}, \pm \frac{4}{5}, \pm \frac{6}{5}, \pm \frac{12}{5}\).
02
Test the possible rational roots
Use synthetic division to test the possible rational roots. Start with \(x = 1\). Synthetic division shows that \(x = 1\) is a root. Perform synthetic division: 1. Coefficients: \(5, 8, -19, -24, 12\)2. Bring down the 5.3. Multiply 5 by 1, add to 8, giving 13.4. Multiply 13 by 1, add to -19, giving -6.5. Multiply -6 by 1, add to -24, giving -30.6. Multiply -30 by 1, add to 12, giving 0.The remainder is 0, thus \(x = 1\) is a root, and \(P(x)\) can be factored as \((x-1)(5x^3 + 13x^2 - 6x -30)\).
03
Further factor the cubic polynomial
Consider the cubic polynomial \(5x^3 + 13x^2 - 6x -30\). Continue applying the Rational Root Theorem to find another root, testing \(x = 2\), for instance. Using synthetic division for \(x = 2\), find that the remainder is 0, which means \(x = 2\) is a root. Perform synthetic division again:1. Coefficients: \(5, 13, -6, -30\)2. Using root \(x = 2\).3. Remainder results in 0, thus dividing, leaving \((5x^2 + 23x + 15)\).Thus, \((5x^3 + 13x^2 - 6x -30)\) factors as \((x-2)(5x^2 + 23x + 15)\).
04
Factor the remaining quadratic
Finally, factor the quadratic \(5x^2 + 23x + 15\). Look for factors of \(5 \times 15 = 75\) that add up to 23. These factors are 3 and 25, realizing it doesn't perfectly work here, try the quadratic formula. The quadratic factors don't provide rational roots, so the quadratic is irreducible over the rationals: \[(5x^2 + 23x + 15) = 0\], roots \(x = \frac{-23 \pm \sqrt{13}}{10}\).Thus, the factorization over the rationals is complete.
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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 process of breaking down a polynomial into simpler polynomials whose product gives the original polynomial. The result is often a set of linear factors—expressions where each factor is of the form \((x - a)\), where \(a\) is a root of the polynomial, or the value where the polynomial equals zero. This process helps to understand the roots of the polynomial and is fundamental in algebra.To factor a polynomial, one can use various techniques:
- Rational Root Theorem: This theorem suggests that every rational root of polynomial \(P(x)\), located in the form \(\frac{p}{q}\), comes from the factors of the constant term divided by the factors of the leading coefficient.
- Synthetic Division: This is a simplified version of polynomial division, dispatched to confirm one potential root at a time. Once a root is confirmed with a remainder of zero, the polynomial can be factored into simpler parts.
- Quadratic Formulas: If a polynomial is reduced to a quadratic form, the quadratic formula can extract complex roots, ensuring that even non-rational roots are factioned properly.
Synthetic Division
Synthetic division is a streamlined method for dividing a polynomial by a binomial of the form \((x-a)\). This method significantly simplifies operations compared to traditional long division, allowing for quick verification of potential roots—especially when paired with the Rational Zeros Theorem.Here's a brief summary of how synthetic division works:
- List all coefficients of the polynomial. For example, in the polynomial \(5x^3 + 13x^2 - 6x - 30\), the coefficients are 5, 13, -6, and -30.
- Pick a possible root, say \(x = 1\), and write it to the left outside of the division bracket.
- Bring down the leading coefficient directly to the bottom row.
- Multiply this number by the chosen root value, place the result under the next coefficient, add them together and continue the process.
- If the last number (remainder) is zero, then the chosen root is indeed a root of the polynomial.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of a quadratic equation, often essential in polynomial factorization when dealing with irreducible quadratic forms. It is primarily useful when the quadratic cannot be factored easily through conventional means, offering a method to find both real and complex roots.The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). The formula to find the roots is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To use the quadratic formula:
- Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation.
- Substitute these values into the formula.
- Solve for \(x\), considering both the \(+\) and \(-\) variations to find two potential zeros.