Chapter 4: Problem 5
Determine if \(g(x)\) is a factor of \(f(x)\) without using synthetic division or long division. $$f(x)=x^{5}-3 x^{3}-2 x^{2} ; \quad g(x)=x-2$$
Short Answer
Expert verified
Answer: Yes, g(x) = x - 2 is a factor of f(x) = x^5 - 3x^3 - 2x^2.
Step by step solution
01
Find the root of g(x)
To find the root of the function \(g(x)=x-2\), set g(x) equal to 0 and solve for x:
$$g(x)=0$$
$$x-2=0$$
$$x=2$$
02
Substitute the root into f(x)
Now, substitute the root x=2 into the function \(f(x)\) to check if \(f(2)=0\):
$$f(x)=x^{5}-3 x^{3}-2 x^{2}$$
$$f(2)=2^{5}-3(2)^{3}-2(2)^{2}$$
03
Simplify f(2)
Evaluate the expression to find the value of f(2):
$$f(2) = 32 - 3(8) - 2(4)$$
$$f(2) = 32 - 24 - 8$$
$$f(2) = 0$$
04
Determine if g(x) is a factor of f(x)
Since \(f(2)=0\), the Factor Theorem tells us that g(x) is indeed a factor of f(x) because the root of g(x) makes f(x) equal to 0. So, $$g(x) = x-2$$ is a factor of $$f(x) = x^{5}-3 x^{3}-2 x^{2}.$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Functions
Polynomial functions are mathematical expressions that involve a sum of powers of variables. They are one of the most fundamental topics in algebra. In simple terms, a polynomial function can take the form of something like \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where each term is a product of a coefficient \(a\) and a variable \(x\) raised to a power.
- Each component, like \(a_nx^n\), is called a "term," and the degree of the polynomial is the highest power of \(x\) with a non-zero coefficient.
- Polynomials can be as simple as a constant number or as complicated as any number of terms added together.
- They can be classified by degree, such as linear (degree 1), quadratic (degree 2), cubic (degree 3), etc.
Root Finding
Root finding is the process of identifying values that make a function equal to zero. In the context of polynomial functions, these values are known as "roots" or "zeros". Solving the equation \(f(x) = 0\) gives us the roots of the polynomial.
- To find the roots of a polynomial, set the polynomial equal to 0 and solve for \(x\).
- There are various methods to identify roots, such as factoring, using the quadratic formula, or more advanced techniques like synthetic division.
- In our original exercise, we found that substituting the root \(x=2\) into \(f(x)\) resulted in a zero output, telling us that \(x=2\) is indeed a root of \(f(x)\).
Algebraic Factoring
Algebraic factoring is an essential process in algebra where a polynomial is expressed as a product of its factors. Factors are polynomials of lower degrees, and this process makes complex equations more manageable.
- Factoring is crucial for solving equations that involve polynomial expressions. It essentially "undoes" the distribution of polynomial terms.
- The Factor Theorem is vital here—if \(f(a) = 0\), then \(x - a\) is a factor of the polynomial \(f(x)\).
- In our exercise, because \(f(2) = 0\), the Factor Theorem confirms that \(x - 2\) is a factor of \(f(x) = x^5 - 3x^3 - 2x^2\).