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Magazine Chapter 15: Problem 17
Find the remaining roots of the given equations using synthetic division, given the roots indicated. $$6 x^{4}+5 x^{3}-15 x^{2}+4=0 \quad\left(r_{1}=-\frac{1}{2}, r_{2}=\frac{2}{3}\right)$$
Short Answer
Expert verified
The remaining roots are \(x = -2\) and \(x = 1\).
Step by step solution
01
Write the Synthetic Division Setup for First Root
For the given root \( r_1 = -\frac{1}{2} \), set up synthetic division. Write \(-\frac{1}{2}\) on the left, and write the coefficients of \(6x^4 + 5x^3 - 15x^2 + 0x + 4\) as \( [6, 5, -15, 0, 4] \).
02
Perform Synthetic Division for First Root
Perform synthetic division using \(-\frac{1}{2}\) with the coefficients:1. Bring down the 6.2. Multiply \(6\) by \(-\frac{1}{2}\) and add to 5 to get 2.3. Multiply \(2\) by \(-\frac{1}{2}\) and add to -15 to get -16.4. Multiply \(-16\) by \(-\frac{1}{2}\) and add to 0 to get 8.5. Multiply \(8\) by \(-\frac{1}{2}\) and add to 4 to get 0 (remainder).
03
Interpret Results of First Synthetic Division
The result of the first synthetic division is \([6, 2, -16, 8] \), indicating the polynomial \(6x^3 + 2x^2 - 16x + 8\). Since the remainder is 0, \(-\frac{1}{2}\) is a valid root.
04
Perform Synthetic Division for Second Root
Now perform synthetic division using the second root \( r_2 = \frac{2}{3} \) with the new coefficients \([6, 2, -16, 8]\):1. Bring down the 6.2. Multiply \(6\) by \(\frac{2}{3}\) and add to 2 to get 6.3. Multiply \(6\) by \(\frac{2}{3}\) and add to -16 to get -12.4. Multiply \(-12\) by \(\frac{2}{3}\) and add to 8 to get 0 (remainder).
05
Interpret Results of Second Synthetic Division
The result of the second synthetic division is \([6, 6, -12] \), indicating the polynomial \(6x^2 + 6x - 12\). Since the remainder is 0, \(\frac{2}{3}\) is a valid root.
06
Solve Quadratic Polynomial
The remaining polynomial from the synthetic division is \(6x^2 + 6x - 12\). Simplify it by dividing by 6, yielding \(x^2 + x - 2 = 0\). Factor the quadratic: \((x + 2)(x - 1) = 0\).
07
Find Remaining Roots
The factored form \((x + 2)(x - 1) = 0\) gives the solutions: \( x = -2 \) and \( x = 1 \). These are the remaining roots of the equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Roots
Finding the roots of a polynomial is like uncovering its secrets! These roots are the values of the variable, usually represented by \( x \), that make the whole polynomial equal zero. For a polynomial of degree \( n \), you can expect up to \( n \) roots, depending on its construction. In our exercise, the polynomial is of degree 4, so we anticipate four roots.
- Given Roots: With roots such as \( r_1 = -\frac{1}{2} \) and \( r_2 = \frac{2}{3} \), we've got a head start. Knowing some roots helps simplify the process.
- Synthetic Division: This technique is used to simplify polynomials, so we can more easily find other roots. When we use a known root in synthetic division, we're essentially dividing the polynomial by \( x - \text{root} \).
- Remainder Zero: If you get a remainder of zero, it confirms that the given number is indeed a root.
Quadratic Equations
Quadratic equations are a special case of polynomials, specifically when the degree is 2. A quadratic is expressed in the standard form \( ax^2 + bx + c = 0 \). In this exercise, after some synthetic division, we're left with the quadratic \(6x^2 + 6x - 12\).
- Simplifying: Always aim to simplify by factoring common terms. When you divide everything by 6, you get \(x^2 + x - 2 = 0\).
- Solving: Factorizing this quadratic into \((x + 2)(x - 1) = 0\) reveals the remaining roots.
Factoring
Factoring is like breaking down a number or expression into products of simpler elements that, when multiplied together, give you the original number or expression. It's a vital tool in solving polynomials.
- Purpose: By expressing a polynomial in factored form, it becomes much easier to find its roots. For the quadratic \(x^2 + x - 2\), we factor it into \((x + 2)(x - 1)\).
- Checking Work: Always ensure the accuracy of factoring by expanding the factors to see if you return to the original expression.
