Problem 106
Question
Factor the expression completely. \(1+8 y^{3}\)
Step-by-Step Solution
Verified Answer
The expression \(1 + 8y^3\) factors to \((1 + 2y)(4y^2 - 2y + 1)\).
1Step 1: Identify the sum of cubes
The expression given is \(1 + 8y^3\). Notice that this expression can be expressed as the sum of cubes, since \(1 = 1^3\) and \(8y^3 = (2y)^3\). We then rewrite the expression as a sum of cubes: \(1^3 + (2y)^3\).
2Step 2: Use the sum of cubes formula
The formula for factoring the sum of cubes \(a^3 + b^3\) is \((a + b)(a^2 - ab + b^2)\). In this case, assign \(a = 1\) and \(b = 2y\). Substitute into the formula: \((1 + 2y)((1)^2 - 1(2y) + (2y)^2).\)
3Step 3: Simplify the expression
Now we simplify the expression using algebraic operations. Perform the calculations inside the compound factor:1. \((1)^2 = 1\)2. \(-1 \times 2y = -2y\)3. \((2y)^2 = 4y^2\)So the expression becomes:\[(1 + 2y)(1 - 2y + 4y^2).\]
4Step 4: Combine the simplified terms
The factored form of the original expression \(1 + 8y^3\) is \((1 + 2y)(4y^2 - 2y + 1)\). Both parts of this product do not factor further, so this is the expression completely factored.
Key Concepts
Sum of CubesPolynomial ExpressionsAlgebraic Formulas
Sum of Cubes
When dealing with algebraic expressions like the one in the exercise, identifying structure can simplify your task. The Sum of Cubes refers to expressions that can be written as the sum of two cube terms. In our case, the expression \(1 + 8y^3\) can be perceived as \(1^3 + (2y)^3\). Recognizing this pattern is key to factoring these types of expressions.
The formula for the sum of cubes, \(a^3 + b^3\), is formulated as:
The formula for the sum of cubes, \(a^3 + b^3\), is formulated as:
- \((a + b)(a^2 - ab + b^2)\)
Polynomial Expressions
Polynomial expressions are algebraic expressions consisting of terms involving variables raised to whole number powers and their coefficients. In the context of this exercise, our polynomial is \(1 + 8y^3\). Here, the powers are manageable, making it easier to apply factoring techniques.
Factoring helps to break down complex polynomial expressions into simpler factors, which can be quite handy in solving equations, simplifying expressions, or finding roots. Each term in a polynomial might not obviously show a factoring pattern unless you can see it as a product of simpler terms or as a well-known identity, like the sum of cubes.
Factoring helps to break down complex polynomial expressions into simpler factors, which can be quite handy in solving equations, simplifying expressions, or finding roots. Each term in a polynomial might not obviously show a factoring pattern unless you can see it as a product of simpler terms or as a well-known identity, like the sum of cubes.
- Identifying patterns or using identities often helps.
- Know common algebraic formulas to factor easily.
Algebraic Formulas
Algebraic formulas are essential tools in simplifying and solving mathematical expressions. They serve as shortcuts to more complex operations, streamlining the process. In our task, we use the formula for the sum of cubes. Recognizing when and where to apply these formulas is integral to solving algebraic problems.
Typical formulas that you should memorize include:
Typical formulas that you should memorize include:
- Sum of Cubes: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
- Difference of Cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
- Perfect Square Trinomials, Quadratic Formulas, and more
Other exercises in this chapter
Problem 105
Clear fractions and solve. $$ \frac{1}{x}+\frac{3 x}{2 x-1}=0 $$
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Rationalize the denominator. $$ \frac{\sqrt{3}-1}{\sqrt{3}+1} $$
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Simplify the expression and write it with rational exponents. Assume that all variables are positive. $$ \frac{1}{2} \sqrt{x}\left(\sqrt{x}+\sqrt[4]{x^{2}}\righ
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Clear fractions and solve. $$ \frac{x}{2 x-5}+\frac{4}{x}=0 $$
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