Problem 15
Question
Find the (real) zeros of the polynomial given. (a) \(f(x)=2 x^{3}+2 x^{2}-12 x\) (b) \(g(x)=2 x^{3}+2 x^{2}+12 x\)
Step-by-Step Solution
Verified Answer
The real zeros of the polynomial \(f(x)=2x^3 + 2x^2 - 12x\) are \(x = 0, 2, -3\). The polynomial \(g(x)=2x^3 + 2x^2 + 12x\) has only one real zero at \(x = 0\).
1Step 1: Identify Common Factors
In both polynomials, \(f(x)\) and \(g(x)\), the common factor is 2x, factoring that out simplifies the polynomial equations to: \[f(x)=2x(x^{2}+x-6)\] and \[g(x)=2x(x^{2}+x+6).\]
2Step 2: Solve for x in \(f(x)\)
Set \(f(x)\) equal to zero and solve for x: \[2x(x^{2}+x-6)=0.\] The solutions are \(x=0\) and also the solutions of the quadratic equation \(x^{2}+x-6=0\), which can be solved by factoring into \((x - 2)(x + 3) = 0\), yielding the solutions \(x = 2\) and \(x = -3\). So the real zeros of the polynomial \(f(x)\) are \(x = 0, 2, -3\).
3Step 3: Solve for x in \(g(x)\)
Set \(g(x)\) equal to zero and solve for x: \[2x(x^{2}+x+6)=0.\] The solutions are \(x=0\) and also the solutions of the quadratic equation \(x^{2}+x+6=0.\) But since the determinant of this equation is negative (\(b^2 - 4ac = 1 - 4*6 < 0)\), it does not have real roots. So, the only real zero of the polynomial \(g(x)\) is \(x = 0\).
Key Concepts
Factoring PolynomialsQuadratic EquationsReal Roots
Factoring Polynomials
Factoring polynomials is a method used to rewrite a polynomial as a product of simpler polynomials. This is a crucial step when trying to find the roots or zeros of a polynomial equation. In simpler terms, you find expressions that multiply together to give you the original polynomial. This step can make solving complex polynomials straightforward.For example, with the polynomial \(f(x) = 2x^3 + 2x^2 - 12x\), the first step is to identify a common factor. Here, it's possible to factor out \(2x\), which simplifies the polynomial to \(2x(x^2 + x - 6)\). By breaking down the polynomial into simpler terms, each factor can be set equal to zero, allowing us to solve for the variable \(x\) more easily.Similarly, for the polynomial \(g(x) = 2x^3 + 2x^2 + 12x\), factoring out \(2x\) yields \(2x(x^2 + x + 6)\). This process highlights the importance of factoring: it simplifies polynomials and sets the stage for solving quadratic equations within them.
Quadratic Equations
Quadratic equations are polynomial equations of degree two, typically in the form \(ax^2 + bx + c = 0\). Solving quadratic equations is often a central problem-solving step when working with higher-degree polynomials after factoring.The quadratic equation in \(f(x) = 2x(x^2 + x - 6)\) after factoring is \(x^2 + x - 6 = 0\). This quadratic can be solved using methods such as factoring, using the quadratic formula, or completing the square. In this case, the quadratic was factored into \((x - 2)(x + 3) = 0)\), giving the solutions \(x = 2\) and \(x = -3\).For \(g(x) = 2x(x^2 + x + 6)\), solving \(x^2 + x + 6 = 0\) presents a different challenge. This equation does not have real solutions because its determinant \((b^2 - 4ac)\) is negative, meaning the equation has complex roots instead of real roots.
Real Roots
Real roots, or real zeros, of a polynomial function are the values of \(x\) where the function equals zero, and the resulting values are real numbers (as opposed to imaginary or complex numbers).In the given exercise, finding real roots begins with the factored form of the polynomial. For \(f(x) = 2x(x^2 + x - 6)\), setting the equation to zero gives two parts to consider: \(2x = 0\), resulting in \(x = 0\); and \(x^2 + x - 6 = 0\), resulting in \(x = 2\) and \(x = -3\) through factoring. These are the real roots of \(f(x)\).For \(g(x) = 2x(x^2 + x + 6)\), setting \(2x = 0\) gives the real root \(x = 0\). However, the quadratic \(x^2 + x + 6 = 0\) has no real roots due to a negative determinant \(1 - 24 \), leading to complex solutions. Thus, \(x = 0\) is the only real root for \(g(x)\). Understanding how to find real roots is critical in determining where the polynomial function crosses the x-axis in a graph, providing valuable insights into its behavior.
Other exercises in this chapter
Problem 14
Suppose that \(f(x)\) is a rational function with zeros at \(x=0\) and \(x=4\), vertical asymptotes at \(x=-2\) and \(x=3\), and a horizontal asymptote at \(y=5
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Find and classify all critical points. \(f(x)=x^{3}+x^{2}+x+1\)
View solution Problem 15
Find and classify all critical points. $$ f(x)=-2 x^{3}+x^{2}+7 $$
View solution Problem 15
The functions that follow in this exercise are not polynomials. We ask you about their range, domain, and graphs with the goal of having you appreciate how nice
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