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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.
Understanding the properties of polynomial functions is crucial for tackling various algebraic operations, including factoring and root finding. When working with higher-degree polynomials, you often look for ways to simplify them, which might involve finding factors—other polynomial expressions that multiply together to give the original polynomial.
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 \(f(x)\).
Finding roots is essential in algebra because it allows us to solve polynomial equations. When a polynomial is factored completely, its roots correspond to the solutions of associated equations. Recognizing and determining roots aids in simplifying complex mathematical problems.
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 \(f(x) = x^5 - 3x^3 - 2x^2\).
Understanding how to factor polynomials helps in numerous applications, from solving algebraic equations to defining the behavior of polynomial functions. Mastering this skill will also aid in finding roots and simplifying expressions across more advanced mathematical contexts.

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