Chapter 12: Problem 34
Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). Find a fourth-degree polynomial function that has zeros \(\sqrt{2},-\sqrt{2}, 1,\) and -1 and a graph that passes through (2,-20).
Short Answer
Expert verified
The polynomial is \(f(x) = -\frac{10}{3}(x^4 - 3x^2 + 2)\).
Step by step solution
01
Identify the factors from the zeros
Since the polynomial has zeros \(\sqrt{2}, -\sqrt{2}, 1\), and \(-1\), we can express the polynomial as a product of its factors based on these zeros: \[(x - \sqrt{2})(x + \sqrt{2})(x - 1)(x + 1).\]
02
Multiply the conjugate pairs
First, we'll multiply the conjugate pairs to simplify the expression:- For \((x - \sqrt{2})(x + \sqrt{2})\), we use the difference of squares formula: \((x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2\).- For \((x - 1)(x + 1)\), we also use the difference of squares formula: \((x - 1)(x + 1) = x^2 - 1\).
03
Multiply the resulting expressions
Now, multiply the results from Step 2: \[(x^2 - 2)(x^2 - 1).\]Expand this product: \[(x^2 - 2)(x^2 - 1) = x^4 - x^2 - 2x^2 + 2 = x^4 - 3x^2 + 2.\]
04
Apply the point condition to determine the leading coefficient
Since the polynomial passes through the point \((2, -20)\), substitute \(x = 2\) and \(f(x) = -20\) into the polynomial \[f(x) = a(x^4 - 3x^2 + 2).\]Calculate: \[a(2^4 - 3(2)^2 + 2) = -20\]\[a(16 - 12 + 2) = -20\]\[a \cdot 6 = -20\]\[a = -\frac{10}{3}.\]
05
Write the polynomial function
Substitute the value of \(a\) back into the polynomial:\[f(x) = -\frac{10}{3}(x^4 - 3x^2 + 2).\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fourth-degree polynomial
A fourth-degree polynomial is a polynomial expression where the highest power of its variable, typically denoted as \(x\), is 4. In general, this kind of polynomial can be expressed in the form:\[a_n x^4 + a_{n-1} x^3 + a_{n-2} x^2 + a_{n-3} x + a_0\]where \(a_n\), \(a_{n-1}\), \(a_{n-2}\), \(a_{n-3}\), and \(a_0\) are constants, with \(a_n eq 0\).
Notably, a fourth-degree polynomial is also known as a quartic polynomial. It can have up to 4 real roots or zeros, which are the values of \(x\) for which the polynomial equals zero.
Notably, a fourth-degree polynomial is also known as a quartic polynomial. It can have up to 4 real roots or zeros, which are the values of \(x\) for which the polynomial equals zero.
- A fourth-degree polynomial graph can take different shapes, depending on the sign and value of the leading coefficient.
- The polynomial can have an upward or downward parabolic beginning and end due to the quartic term \(x^4\).
- It may have local maxima and minima, based on the arrangement of its zeros.
Zeros of a polynomial
The zeros of a polynomial, also known as roots, are the values of \(x\) that make the polynomial equal to zero. In other words, they are the solutions to the polynomial equation \(f(x) = 0\).
For a fourth-degree polynomial, it can have up to four zeros. These zeros are crucial as they help in forming the polynomial using its factorized form.For instance, if a polynomial has zeros at \(\sqrt{2}, -\sqrt{2}, 1,\) and \(-1\), the factorized form would look like:- \((x - \sqrt{2})(x + \sqrt{2})(x - 1)(x + 1)\)
For a fourth-degree polynomial, it can have up to four zeros. These zeros are crucial as they help in forming the polynomial using its factorized form.For instance, if a polynomial has zeros at \(\sqrt{2}, -\sqrt{2}, 1,\) and \(-1\), the factorized form would look like:- \((x - \sqrt{2})(x + \sqrt{2})(x - 1)(x + 1)\)
- Zeros help identify the points where the graph of the polynomial crosses the x-axis.
- Real zeros are visible as x-intercepts on the graph, while complex zeros come in conjugate pairs and do not appear as x-intercepts for real-valued polynomials.
Difference of squares
The difference of squares is a specific algebraic formula used to simplify expressions in the form of \((a - b)(a + b)\). The generalized identity is:\[(a - b)(a + b) = a^2 - b^2\]This formula is highly useful when dealing with conjugate pairs, such as \((x - \sqrt{2})(x + \sqrt{2})\) or \((x - 1)(x + 1)\).
- For \((x - \sqrt{2})(x + \sqrt{2})\), using the difference of squares gives us \(x^2 - 2\).
- Similarly, \((x - 1)(x + 1)\) simplifies to \(x^2 - 1\).
Leading coefficient
The leading coefficient of a polynomial is the coefficient attached to the term with the highest degree. In a fourth-degree polynomial, this would be the coefficient of the \(x^4\) term. It is a critical part of the polynomial's structure as it influences the graph's end behavior.Here are some key points about the leading coefficient:- Determines whether the ends of the polynomial curve rise or fall as \(x\) approaches infinity or negative infinity.- The sign of the leading coefficient indicates if the graph is pointing upwards or downwards at both ends (in the case of a quartic, both ends either rise or fall together when non-complex).- When scaling the polynomial to pass through a specific point, like (2, -20), this coefficient adjusts to fit the curve properly. For example, if a polynomial must pass through the point (2, -20), the leading coefficient can be determined by substituting into the expanded polynomial and solving for it. After finding the zeros and simplifying, the leading coefficient \(a\) was calculated as \(-\frac{10}{3}\) in this case, ensuring the polynomial fits the condition given in the problem.