Chapter 3: Problem 39
Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$P(x)=x^{4}-3 x^{2}-4$$
Short Answer
Expert verified
The zeros of the polynomial are \(x = -2\) and \(x = 2\).
Step by step solution
01
Recognize Polynomial Structure
Observe that the given polynomial can be treated as a quadratic in terms of \(x^2\), i.e., \(P(x) = (x^2)^2 - 3(x^2) - 4\). This allows us to perform a substitution to simplify it.
02
Substitute and Simplify
Let \(u = x^2\). Then, the polynomial becomes \(u^2 - 3u - 4\). This is a quadratic equation in \(u\), which we can factor.
03
Factor the Quadratic
Factor the quadratic \(u^2 - 3u - 4\) by finding two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(+1\). So, the factorization is \((u - 4)(u + 1)\).
04
Substitute Back
Replace \(u\) with \(x^2\) in the factorization to get \((x^2 - 4)(x^2 + 1)\).
05
Further Factor
Notice \(x^2 - 4\) is a difference of squares and can be further factored as \((x - 2)(x + 2)\). The term \(x^2 + 1\) does not factor over the real numbers.
06
Identify Zeros
To find the zeros of \(P(x)\), solve \((x - 2)(x + 2) = 0\). Thus, \(x = 2\) and \(x = -2\) are the zeros. The equation \(x^2 + 1 = 0\) gives no real solutions as it leads to a negative inside the square root.
07
Sketch the Graph
Plot the polynomial with zeros at \(x = -2\) and \(x = 2\). Since the degree is 4, the arms of the graph extend upwards in both directions. The polynomial does not cross at \((x^2 + 1)\) because there are no real roots for this factor.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Zeros of Polynomials
Understanding the zeros of a polynomial is crucial because these values dictate where the graph of the polynomial intersects the x-axis. Zeros, also known as roots, solutions, or x-intercepts, satisfy the equation \(P(x) = 0\). In our problem, the polynomial \(P(x) = x^4 - 3x^2 - 4\) was initially transformed into a simpler quadratic form. By letting \(u = x^2\), we found the zeros of \(u^2 - 3u - 4\) as \(u = 4\) and \(u = -1\). Subsequently, replacing \(u\) back with \(x^2\), we determined the zeros of the actual polynomial.
- Solving \(x^2 - 4 = 0\) led us to find \(x = \pm2\) as real zeros.
- The factor \(x^2 + 1 = 0\) yields no real solutions because solving this would entail finding the square root of \(-1\), which does not exist among real numbers.Zeros are pivotal for graphing, and understanding their absence in certain parts can explain why graphs don't touch or cross the x-axis at specific intervals.
- Solving \(x^2 - 4 = 0\) led us to find \(x = \pm2\) as real zeros.
- The factor \(x^2 + 1 = 0\) yields no real solutions because solving this would entail finding the square root of \(-1\), which does not exist among real numbers.Zeros are pivotal for graphing, and understanding their absence in certain parts can explain why graphs don't touch or cross the x-axis at specific intervals.
Quadratic Equations
Quadratic equations, which take the form \(ax^2 + bx + c = 0\), are fundamental in polynomial factoring. In this discussion, we convert \(P(x) = x^4 - 3x^2 - 4\) into a quadratic-like structure by using a substitution of \(u = x^2\). This allows the equation to resemble a standard quadratic form, \(u^2 - 3u - 4\).
The process of factoring a quadratic requires finding two numbers that multiply to the constant term, \(-4\), and add up to the coefficient of the linear term, \(-3\). These numbers are \(-4\) and \(+1\), leading to the factorization \((u - 4)(u + 1)\).
Once factored, it becomes straightforward to solve for \(u\), reveal its equivalent original variable \(x\), and obtain real solutions.
By understanding how quadratic equations are factored and solved, we decode a vital aspect of algebra that aids in graph interpretation and solving higher-degree polynomials.
The process of factoring a quadratic requires finding two numbers that multiply to the constant term, \(-4\), and add up to the coefficient of the linear term, \(-3\). These numbers are \(-4\) and \(+1\), leading to the factorization \((u - 4)(u + 1)\).
Once factored, it becomes straightforward to solve for \(u\), reveal its equivalent original variable \(x\), and obtain real solutions.
By understanding how quadratic equations are factored and solved, we decode a vital aspect of algebra that aids in graph interpretation and solving higher-degree polynomials.
Graphing Polynomials
Graphing polynomials involves plotting points that symbolize solutions—or zeros—on a coordinate plane. For \(P(x) = x^4 - 3x^2 - 4\), we identified the zeros \(x = -2\) and \(x = 2\). These zeros display where the graph touches or crosses the x-axis.
When sketching polynomial graphs:
When sketching polynomial graphs:
- Note the degree of the polynomial, which determines the graph's end behavior. Here, the degree is 4, meaning the graph extends upwards at both ends.
- Identify all real zeros, since they are the points where the curve intersects the x-axis.
- Understand that non-real zeros, such as those from \(x^2 + 1 = 0\), imply no intersection at certain parts of the graph, as these do not contribute additional x-intercepts.