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Magazine Chapter 3: Problem 28
Find all solutions of the equation. $$6 x^{4}+5 x^{3}-17 x^{2}-6 x=0$$
Short Answer
Expert verified
The solutions are \( x = 0, x = -\frac{1}{2}, x = 1.5, \text{and } x = -\frac{4}{3} \).
Step by step solution
01
Factor out the greatest common factor
The given polynomial equation is \( 6x^4 + 5x^3 - 17x^2 - 6x = 0 \). First, we notice that x is a common factor in all terms. Factor x out: \( x(6x^3 + 5x^2 - 17x - 6) = 0 \). This immediately gives us one solution: \( x = 0 \).
02
Apply the Rational Root Theorem to the cubic
With the cubic polynomial \( 6x^3 + 5x^2 - 17x - 6 = 0 \), apply the Rational Root Theorem to identify possible rational roots as fractions of the form \( \frac{p}{q} \) where \( p \) divides the constant term (-6) and \( q \) divides the leading coefficient (6). The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2} \).
03
Test potential rational roots
Substitute each potential rational root into the polynomial equation \( 6x^3 + 5x^2 - 17x - 6 \) and determine which values make the polynomial equal to zero. After testing, you find that \( x = -\frac{1}{2} \) is a root.
04
Use synthetic division to factor the cubic
Having determined that \( x = -\frac{1}{2} \) is a root, use synthetic division to divide the cubic polynomial \( 6x^3 + 5x^2 - 17x - 6 \) by \( (x + \frac{1}{2}) \). This gives you a quadratic polynomial quotient: \( 6x^2 - x - 12 \).
05
Solve the quadratic equation
The quadratic equation \( 6x^2 - x - 12 = 0 \) can be solved using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6, b = -1, \text{ and } c = -12 \). This yields the solutions \( x = 1.5 \) and \( x = -\frac{4}{3} \).
06
Compile all solutions
The solutions from factoring and solving the polynomial are \( x = 0, x = -\frac{1}{2}, x = 1.5, \text{and } x = -\frac{4}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorization
Factorization is the process of breaking down a complex expression into a product of simpler factors. When dealing with polynomial equations, it's often useful to start by identifying a common factor in each term. In the original equation, \(6x^4 + 5x^3 - 17x^2 - 6x = 0\), it's evident that \(x\) is a common factor across all terms. Factoring \(x\) out simplifies the polynomial significantly.
- We rewrote the polynomial as \(x(6x^3 + 5x^2 - 17x - 6) = 0\). This immediately gives a simple solution: \(x = 0\).
- After factoring out the greatest common factor, we are left with the cubic polynomial \(6x^3 + 5x^2 - 17x - 6\). This is a more manageable equation to work with moving forward in the problem-solving process.
Rational Root Theorem
The Rational Root Theorem is a powerful tool for finding rational solutions to polynomial equations. It states that for any polynomial with integer coefficients, any potential rational roots, \(\frac{p}{q}\), must have \(p\) as a factor of the constant term of the polynomial and \(q\) as a factor of the leading coefficient.
Considering the polynomial \(6x^3 + 5x^2 - 17x - 6\), we see:
Considering the polynomial \(6x^3 + 5x^2 - 17x - 6\), we see:
- The constant term is \(-6\).
- The leading coefficient is \(6\).
- This means the possible rational roots are \(\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}\).
Synthetic Division
Synthetic division is a quick method for dividing a polynomial by a binomial of the form \(x - c\). This technique is particularly helpful once a root has been identified using methods like the Rational Root Theorem.
In our example, we confirmed that \(x = -\frac{1}{2}\) is a root of the polynomial \(6x^3 + 5x^2 - 17x - 6\). Using synthetic division, we divide by \((x + \frac{1}{2})\) to reduce the cubic polynomial into a quadratic form.
In our example, we confirmed that \(x = -\frac{1}{2}\) is a root of the polynomial \(6x^3 + 5x^2 - 17x - 6\). Using synthetic division, we divide by \((x + \frac{1}{2})\) to reduce the cubic polynomial into a quadratic form.
- The resulting quotient from the division was \(6x^2 - x - 12\).
- This simplifies our task as solving a quadratic is simpler than a cubic.
Quadratic Formula
The quadratic formula is a well-known method for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let's break down the solution process using this formula for the quadratic equation derived from synthetic division: \(6x^2 - x - 12 = 0\).
Let's break down the solution process using this formula for the quadratic equation derived from synthetic division: \(6x^2 - x - 12 = 0\).
- Here, \(a = 6\), \(b = -1\), and \(c = -12\).
- Substitute these values into the quadratic formula.
- Calculate the discriminant \(b^2 - 4ac\): \((-1)^2 - 4(6)(-12)\).
- Solve it step-by-step to find the roots: \(x = 1.5\) and \(x = -\frac{4}{3}\).
