Problem 11
Question
A polynomial \(P\) is given. (a) Find all zeros of \(P\), real and complex. (b) Factor \(P\) completely. $$P(x)=x^{4}-16$$
Step-by-Step Solution
Verified Answer
Zeros are \(x = 2, -2, 2i, -2i\); factored form is \((x - 2)(x + 2)(x - 2i)(x + 2i)\).
1Step 1: Identify the Type of Polynomial
The polynomial given is \(P(x) = x^4 - 16\). This is a difference of squares since it can be expressed as \(x^4 - 4^2\).
2Step 2: Apply the Difference of Squares
Recall that the difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\). Here, \(a = x^2\) and \(b = 4\), so \(P(x) = (x^2 - 4)(x^2 + 4)\).
3Step 3: Factor Further the Quadratic Expression
The expression \(x^2 - 4\) is another difference of squares: \(x^2 - 2^2\), which factors to \((x - 2)(x + 2)\). So far we have \((x - 2)(x + 2)(x^2 + 4)\).
4Step 4: Solve the Remaining Quadratic
The quadratic \(x^2 + 4 = 0\) can be rewritten as \(x^2 = -4\). Solving for \(x\), take the square root of both sides to get \(x = \pm 2i\).
5Step 5: List all Zeros of the Polynomial
The zeros of \(P(x)\) are the solutions of the factors. From \((x - 2)(x + 2)\), we have zeros at \(x = 2\) and \(x = -2\). From \(x^2 + 4\), we have zeros at \(x = 2i\) and \(x = -2i\).
6Step 6: Write the Completely Factored Form of P(x)
After identifying all zeros and factors, the completely factored form of \(P(x)\) is \((x - 2)(x + 2)(x - 2i)(x + 2i)\).
Key Concepts
Complex NumbersDifference of SquaresQuadratic Expressions
Complex Numbers
Complex numbers are essential in understanding solutions that seem impossible when dealing with real numbers alone. A complex number is written in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part. The imaginary unit \(i\) is defined by the property \(i^2 = -1\). This allows us to extend our number system to accommodate square roots of negative numbers.
In the context of the given polynomial \(P(x) = x^4 - 16\), we encounter complex numbers when we factor the expression \(x^2 + 4\). Solving \(x^2 + 4 = 0\), we get \(x^2 = -4\). Here is where \(i\) comes into play, as we solve \(x = \pm 2i\) to find complex roots.
Understanding complex numbers is fundamental for solving quadratic equations with negative discriminants and plays a critical role in polynomial factorization.
In the context of the given polynomial \(P(x) = x^4 - 16\), we encounter complex numbers when we factor the expression \(x^2 + 4\). Solving \(x^2 + 4 = 0\), we get \(x^2 = -4\). Here is where \(i\) comes into play, as we solve \(x = \pm 2i\) to find complex roots.
Understanding complex numbers is fundamental for solving quadratic equations with negative discriminants and plays a critical role in polynomial factorization.
Difference of Squares
The difference of squares is a straightforward algebraic identity: \(a^2 - b^2 = (a-b)(a+b)\). Identifying and applying this formula can simplify the factorization of polynomials drastically.
For the given polynomial \(P(x) = x^4 - 16\), recognizing it as a difference of squares allows us to factor it into \((x^2 - 4)(x^2 + 4)\). This is possible because it can be rewritten as \((x^2)^2 - 4^2\), where \(a = x^2\) and \(b = 4\).
Using the difference of squares not only simplifies polynomials like \(x^4 - 16\) but also helps in recognizing patterns that can be factored further into simple, linear factors or other familiar quadratics.
For the given polynomial \(P(x) = x^4 - 16\), recognizing it as a difference of squares allows us to factor it into \((x^2 - 4)(x^2 + 4)\). This is possible because it can be rewritten as \((x^2)^2 - 4^2\), where \(a = x^2\) and \(b = 4\).
Using the difference of squares not only simplifies polynomials like \(x^4 - 16\) but also helps in recognizing patterns that can be factored further into simple, linear factors or other familiar quadratics.
Quadratic Expressions
Quadratic expressions take the form \(ax^2 + bx + c\). In polynomial factorization, breaking down these expressions into simpler linear factors is a common task.
Looking back at the step where we factored \(x^2 - 4\), we recognized it as a difference of squares: \(x^2 - 2^2\), which factors further into \((x - 2)(x + 2)\). This demonstrates how a quadratic expression can be decomposed using basic algebraic identities.
Quadratic expressions are everywhere in algebra, and mastering them enables students to factor polynomials efficiently, solve equations, and find zeros of more complicated expressions. In our example, tackling \(x^2 + 4 = 0\) presented a quadratic with real coefficients but complex roots, emphasizing the connection between quadratics and complex numbers.
Looking back at the step where we factored \(x^2 - 4\), we recognized it as a difference of squares: \(x^2 - 2^2\), which factors further into \((x - 2)(x + 2)\). This demonstrates how a quadratic expression can be decomposed using basic algebraic identities.
Quadratic expressions are everywhere in algebra, and mastering them enables students to factor polynomials efficiently, solve equations, and find zeros of more complicated expressions. In our example, tackling \(x^2 + 4 = 0\) presented a quadratic with real coefficients but complex roots, emphasizing the connection between quadratics and complex numbers.
Other exercises in this chapter
Problem 10
A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x\) - and \(y\) -intercept(s). (c) Sketch its
View solution Problem 11
Find the \(x\) - and \(y\) -intercepts of the rational function. $$r(x)=\frac{x-1}{x+4}$$
View solution Problem 11
Find the real and imaginary parts of the complex number. $$-\frac{2}{3} i$$
View solution Problem 11
Two polynomials \(P\) and \(D\) are given. Use either synthetic or long division to divide \(P(x)\) by \(D(x),\) and express the quotient \(P(x) / D(x)\) in the
View solution