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Magazine Chapter 11: Problem 16
For each polynomial function, one zero is given. Find all others. \(f(x)=x^{3}+2 x^{2}+32 x-80 ; \quad 2\)
Short Answer
Expert verified
The zeros are 2, -2 + 6i, and -2 - 6i.
Step by step solution
01
Verify the given zero
Check if the given zero, 2, satisfies the polynomial function by substituting it into the equation. So, substitute 2 for x in the polynomial:\(f(2) = (2)^3 + 2(2)^2 + 32(2) - 80\)Calculate the value:\(f(2) = 8 + 8 + 64 - 80 = 0\)Since \(f(2) = 0\), 2 is indeed a zero of the polynomial.
02
Perform polynomial division
Divide the polynomial \(f(x) = x^{3}+2 x^{2}+32 x-80\) by \(x - 2\) using synthetic division. The synthetic division setup is as follows:Coefficients: \([1, 2, 32, -80]\)The synthetic division process:\begin{array}{r|rrrr} 2 & 1 & 2 & 32 & -80 \ \text{} & \text{} & 2 & 8 & 80 \text{resulting array} & 1 & 4 & 40 & 0 \text{final result:} & 1 & 4 & 40 \text{remainder} & 0 \text{Thus, quotient: } & x^2 + 4x + 40division result: \(x^{2} + 4x + 40\)
03
Find the zeros of the quotient polynomial
Solve the quadratic equation \(x^{2} + 4x + 40 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).Here, \(a = 1\), \(b = 4\), and \(c = 40\). Substitute these values in:\[x = \frac{-4 \pm \sqrt{4^2 - 4(1)(40)}}{2(1)}\]\[x = \frac{-4 \pm \sqrt{16 - 160}}{2}\]\[x = \frac{-4 \pm \sqrt{-144}}{2}\]\[x = \frac{-4 \pm 12i}{2}\]\[x = -2 \pm 6i\]The solutions are \(x = -2 + 6i\) and \(x = -2 - 6i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
synthetic division
Synthetic division is a simplified form of polynomial division, especially useful when dividing by linear expressions of the form \(x - c\). It's much quicker and allows for less room for error compared to long division.
To perform synthetic division, follow these steps:
To perform synthetic division, follow these steps:
- Identify the coefficients of the polynomial. For example, the polynomial \(x^3 + 2x^2 + 32x -80\) has coefficients [1, 2, 32, -80].
- Write the zero of the divisor. Here, it鈥檚 2 for the divisor \(x-2\).
- Set up the synthetic division table. Place the zero at the leftmost position and the coefficients to its right. It looks like this:
\(\begin{array}{r|rrrr} 2 & 1 & 2 & 32 & -80 \ \text{} & \text{} & 2 & 8 & 80 \ \text{resulting array} & 1 & 4 & 40 & 0 \ \text{final result:} & 1 & 4 & 40\ \text{remainder} & 0\ \text{Thus, quotient: }\) \(x^2 + 4x + 40\)
- The first coefficient (1) is brought down as it is.
- Multiply it by the zero and place the result under the next coefficient (2).
- Continue the process: add, multiply, and bring down until all coefficients are dealt with.
quadratic formula
The quadratic formula is a powerful tool used to find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). It looks like this:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The quadratic formula helps when factoring is difficult or impossible. Here's a step-by-step guide:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The quadratic formula helps when factoring is difficult or impossible. Here's a step-by-step guide:
- Identify coefficients a, b, and c. For \(x^2 + 4x + 40 = 0\), we have \(a = 1, b = 4,\ and \c = 40\).
- Substitute these values into the formula:
\[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(40)}}{2(1)} \] - Calculate under the square root (the discriminant):
\ 4^2 - 4\cdot1\cdot40 = 16 - 160 = -144 \. - The discriminant is negative, leading to complex solutions.
Thus, \[ x = \frac{-4 \pm \sqrt{-144}}{2} \] - Simplify the square root of the negative number:
- \(\sqrt{-144} = 12i \)
- \( x = -2 + 6i \) and \( x = -2 - 6i \).
complex numbers
Complex numbers add a new dimension to solving polynomial equations. They are numbers of the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, defined as \(\sqrt{-1}\).
In our case, the quadratic formula revealed complex roots for the polynomial \ x^2 + 4x + 40 = 0\:\(x = -2 + 6i\) \(x = -2 - 6i\)
In our case, the quadratic formula revealed complex roots for the polynomial \ x^2 + 4x + 40 = 0\:
- These roots cannot be real numbers because they rely on the square root of a negative number.
- \(i\) is the imaginary unit, where \( i = \sqrt{-1} \).
- A complex conjugate is a pair like \(-2 + 6i\) and \(-2 - 6i\). When multiplied, conjugates eliminate the imaginary part.
- Complex numbers follow standard arithmetic rules, but remember that \(i^2 = -1\).
Essential points:
