Problem 28
Question
Factor the polynomial. $$x^{8}-16$$
Step-by-Step Solution
Verified Answer
The polynomial factors as \((x^4 + 4)(x^2 + 2)(x^2 - 2)\).
1Step 1: Recognize the Difference of Squares
Recognize that the polynomial \(x^8 - 16\) can be written as a difference of squares. The expression is equivalent to \((x^4)^2 - 4^2\). This allows us to factor the polynomial using the difference of squares formula, \(a^2 - b^2 = (a + b)(a - b)\).
2Step 2: Apply the Difference of Squares Formula
Apply the difference of squares formula to \( (x^4)^2 - 4^2 \), which results in \((x^4 + 4)(x^4 - 4)\).
3Step 3: Further Factor the Polynomial
Notice that \(x^4 - 4\) is another difference of squares, since it can be written as \((x^2)^2 - 2^2\). Therefore, apply the difference of squares formula again to get \((x^2 + 2)(x^2 - 2)\).
4Step 4: Write the Final Factored Form
Combine all the factors. The polynomial \(x^8 - 16\) is factored completely as \((x^4 + 4)(x^2 + 2)(x^2 - 2)\). Note that \(x^4 + 4\) does not factor further over the reals.
Key Concepts
Difference of SquaresPolynomial FactoringAlgebraic Expressions
Difference of Squares
The concept of the Difference of Squares is a powerful algebraic tool that simplifies the factoring of certain polynomials. It's based on the identity:
In the context of the polynomial \(x^8 - 16\), we first recognized it as \((x^4)^2 - 4^2\). Identifying such expressions is key when using the difference of squares technique.
\((x^4)^2 - 4^2\) can be decomposed using the formula, resulting in \((x^4 + 4)(x^4 - 4)\).
This ability to break down complex expressions into simpler factors is what makes the difference of squares method an essential technique in algebra.
- \(a^2 - b^2 = (a + b)(a - b)\)
In the context of the polynomial \(x^8 - 16\), we first recognized it as \((x^4)^2 - 4^2\). Identifying such expressions is key when using the difference of squares technique.
\((x^4)^2 - 4^2\) can be decomposed using the formula, resulting in \((x^4 + 4)(x^4 - 4)\).
This ability to break down complex expressions into simpler factors is what makes the difference of squares method an essential technique in algebra.
Polynomial Factoring
Polynomial factoring is the process of expressing a polynomial as a product of simpler polynomials. This is crucial in solving polynomial equations, simplifying expressions, and understanding polynomial behavior.
Different methods exist for polynomial factoring, and choosing the appropriate one depends on the polynomial's form.
For \(x^8 - 16\), after using the difference of squares to get \((x^4 + 4)(x^4 - 4)\), further analysis of \(x^4 - 4\) reveals another layer of factoring potential.
Different methods exist for polynomial factoring, and choosing the appropriate one depends on the polynomial's form.
For \(x^8 - 16\), after using the difference of squares to get \((x^4 + 4)(x^4 - 4)\), further analysis of \(x^4 - 4\) reveals another layer of factoring potential.
- We rewrite \(x^4 - 4\) as \((x^2)^2 - (2)^2\), which is again a difference of squares.
- Applying the same method gives us \((x^2 + 2)(x^2 - 2)\).
Algebraic Expressions
Understanding algebraic expressions is fundamental to mastering algebraic manipulations. An algebraic expression consists of variables, constants, and operational symbols that together represent a mathematical reality.
Expressions can often be simplified, factored, or expanded depending on what is needed.
With practice, identifying these can become intuitive, making problem-solving much more straightforward. Always remember when dealing with algebraic expressions to look for patterns or special forms such as the difference of squares.
Expressions can often be simplified, factored, or expanded depending on what is needed.
- For example, in the expression \(x^8 - 16\), we began with recognizing its structure as a difference of squares, which is a particular type of algebraic expression.
- This recognition allowed us to apply factoring techniques efficiently.
With practice, identifying these can become intuitive, making problem-solving much more straightforward. Always remember when dealing with algebraic expressions to look for patterns or special forms such as the difference of squares.
Other exercises in this chapter
Problem 28
Simplify. $$\left(2 x^{2} y^{-5}\right)\left(6 x^{-3} y\right)\left(\frac{1}{3} x^{-1} y^{3}\right)$$
View solution Problem 28
Factor the polynomial. $$x^{4}-8 x^{3}+16 x^{2}$$
View solution Problem 28
Write the expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$(3-2 i)^{3}$$
View solution Problem 29
Solve the equation. \(9 x^{3}-18 x^{2}-4 x+8=0\)
View solution