Problem 27
Question
Complete the following. (a) Find all zeros of \(f(x)\) (b) Write the complete factored form of \(f(x)\) $$ f(x)=x^{4}+5 x^{2}+4 $$
Step-by-Step Solution
Verified Answer
The zeros are \( i, -i, 2i, -2i \); the factored form is \((x-i)(x+i)(x-2i)(x+2i)\).
1Step 1: Substitution to Simplify
Let us make a substitution to simplify the problem: set \( y = x^2 \). Thus, the function becomes a quadratic in terms of \( y \):\[f(x) = x^4 + 5x^2 + 4 = y^2 + 5y + 4\]We now need to find the zeros of this quadratic expression.
2Step 2: Factor the Quadratic Expression
Factor the quadratic expression \( y^2 + 5y + 4 \). We look for two numbers that multiply to 4 and add to 5. The numbers 1 and 4 fit this requirement:\[y^2 + 5y + 4 = (y + 1)(y + 4)\]Substituting back for \( y \), we get:\[(x^2 + 1)(x^2 + 4)\]This corresponds to th factors in terms of \( x \).
3Step 3: Find the Zeros of Each Factor
Solve for the zeros of each factor separately:1. \( x^2 + 1 = 0 \)\[x^2 = -1 \Rightarrow x = \pm i\]2. \( x^2 + 4 = 0 \)\[x^2 = -4 \Rightarrow x = \pm 2i\]Thus, the zeros of \( f(x) \) are \( x = i, -i, 2i, -2i \).
4Step 4: Write the Complete Factored Form
Using the zeros, write the complete factored form of the function:\[f(x) = (x - i)(x + i)(x - 2i)(x + 2i)\]This represents the factored form of \( f(x) \) using complex numbers.
Key Concepts
Factored FormPolynomial EquationsComplex Numbers
Factored Form
In algebra, expressing a polynomial in factored form means breaking it down into a product of simpler polynomials or numbers. This makes it much easier to analyze and solve equations. For the function given, we start by rewriting it as a quadratic expression using a substitution method.
By substituting, we turn the quartic (degree four) polynomial into a quadratic (degree two) in terms of another variable. Once factored, this can reveal useful information about the roots or solutions of the polynomial.
It's like uncovering the hidden parts of a larger equation, allowing us to see what the zeros are and how they multiply or combine together. In the provided solution, notice the zeros given are complex numbers—showing the importance of having both real and imaginary parts accounted for when dealing with polynomials.
By substituting, we turn the quartic (degree four) polynomial into a quadratic (degree two) in terms of another variable. Once factored, this can reveal useful information about the roots or solutions of the polynomial.
It's like uncovering the hidden parts of a larger equation, allowing us to see what the zeros are and how they multiply or combine together. In the provided solution, notice the zeros given are complex numbers—showing the importance of having both real and imaginary parts accounted for when dealing with polynomials.
Polynomial Equations
Polynomial equations are expressions that outlay the relationship between variables and coefficients involving only non-negative integer powers of the variable. The polynomial equation in the exercise is given by:
\[f(x) = x^4 + 5x^2 + 4\]
This is a quartic polynomial because the highest exponent is 4. Solving polynomial equations often requires finding the roots or zeros of the equation, which are the values of the variable that make the expression equal zero.
To tackle this polynomial, the method used is substitution, changing the form temporarily into something more manageable, such as a quadratic. The equation becomes:
\[y^2 + 5y + 4\]
Factoring this quadratic expression gives the potential solutions which can be further expanded to find solutions in the original variable. This method is often preferable over direct factoring of complex polynomials, making it simpler to work with.
\[f(x) = x^4 + 5x^2 + 4\]
This is a quartic polynomial because the highest exponent is 4. Solving polynomial equations often requires finding the roots or zeros of the equation, which are the values of the variable that make the expression equal zero.
To tackle this polynomial, the method used is substitution, changing the form temporarily into something more manageable, such as a quadratic. The equation becomes:
\[y^2 + 5y + 4\]
Factoring this quadratic expression gives the potential solutions which can be further expanded to find solutions in the original variable. This method is often preferable over direct factoring of complex polynomials, making it simpler to work with.
Complex Numbers
Complex numbers arise naturally when solving polynomial equations, especially when the solutions involve square roots of negative numbers.
A complex number is written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. Imaginary numbers themselves are expressed with \(i\), which is the square root of -1.
In our solution, when finding the zeros of factors like \(x^2 + 1\) and \(x^2 + 4\), we encounter imaginary numbers:
Complex numbers play a critical role in ensuring equations with no real solution can still be fully solved. They extend the idea of number systems, helping algebra solve diverse and complicated problems by adding a layer of depth and functionality.
A complex number is written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. Imaginary numbers themselves are expressed with \(i\), which is the square root of -1.
In our solution, when finding the zeros of factors like \(x^2 + 1\) and \(x^2 + 4\), we encounter imaginary numbers:
- \(x = \pm i\)
- \(x = \pm 2i\)
Complex numbers play a critical role in ensuring equations with no real solution can still be fully solved. They extend the idea of number systems, helping algebra solve diverse and complicated problems by adding a layer of depth and functionality.
Other exercises in this chapter
Problem 26
Find all real solutions. Check your results. $$ \frac{1}{x-3}+1=\frac{6}{x^{2}-9} $$
View solution Problem 27
Divide the expression. $$\frac{8 x^{3}+10 x^{2}-12 x-15}{2 x^{2}-3}$$
View solution Problem 27
Use positive exponents to rewrite. $$ \sqrt{y \cdot \sqrt{y}} $$
View solution Problem 27
Find all real solutions. Check your results. $$ \frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x^{2}-1} $$
View solution