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Chapter 6: Problem 29

Factor completely. $$ 1+8 d^{3}+16 d^{6} $$

Short Answer

Expert verified
The factored form is: = [1 + (2d + (2d^2)]

Step by step solution

01

Identify common factors

Observe the given expression: 1 + 8d^3 + 16d^6. Notice that all terms can be written as powers of 2. Rewrite the expression to see this more clearly: 1(=1^2), (2d)^3, (4d^2)^3.
02

Rewrite the expression

Re-write each term in exponent form to make common factors more evident: \[1 + 8d^3 + 16d^6 = 1 + (2d)^3 + (4d^2)^3\]
03

Use substitution for a simpler form

To simplify, let z = 2d. Rewrite the expression with z substitution: \[1 + z^3 + (z^2)^3 = 1 + z^3 + z^6\]
04

Recognize the sum of cubes

Identify the structure of the expression as a sum of cubes and re-group them: \[(1 + z^3 + z^6) = (1 + z^3 + (z^2)^3)\]
05

Apply the sum formula for cubes once more

Finally, factor the expression using sum of cubes formula: \[(1 + z^3 + z^6) = [1 + (z + (z^2)]\]
06

Substitute back to original variable

Returning to the original variable 2d, we have: \[1 + (2d)^6 + (2d)^3\]. This simplifies using step 5 to:\[= [1 + (2d + (2d^2)]\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Cubes
The sum of cubes is a fundamental concept in polynomial factorization. The general form is represented as: \textcolor{blue}{\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). To factor an expression using the sum of cubes formula, it's essential to identify the terms that are perfect cubes. For the given expression, we notice that terms like \(8d^3\) and \(16d^6\) are perfect cubes because:
  • \(8d^3 = (2d)^3\)
  • \(16d^6 = (4d^2)^3\)
Rewriting these terms allows us to recognize the structure of the sum of cubes.
Algebraic Expressions
An algebraic expression is a combination of constants, variables, and algebraic operations such as addition, subtraction, multiplication, and division. In our exercise, the given expression is \(1 + 8d^3 + 16d^6\).
The skill of rewriting such expressions into a more recognizable form helps in simplifying complex problems. Here, we rewrote \(8d^3\) as \((2d)^3\) and \(16d^6\) as \((4d^2)^3\) to identify common factors. Using substitution \(z = 2d\) simplifies the expression further, making manipulation easier and more intuitive.
Polynomial Factorization
Polynomial factorization involves expressing a polynomial as a product of its simpler polynomials. Look at the last factored form: \(1 + (2d + (2d^2)\). Polynomial factorization works by:
  • Identifying and factoring out any common factors.
  • Using special formulas (like the sum of cubes) to split the polynomial into simpler parts.

For comprehensive understanding, factorizing a polynomial leverages algebraic manipulation, formulas and sometimes, substitution to simplify expressions into products of their factors.

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Most popular questions from this chapter

Factor completely. Remember to look first for a common factor. If a polynomial is prime, state this. $$ x^{2}-6 x+9-y^{2} $$

Review multiplying binomials using FOIL $$ (x-2)(x+7) $$

Solve. Round any irrational solutions to the nearest thousandth. $$ a^{2}=\frac{1}{25} $$

Find the zeros of each function. $$ f(x)=x^{2}-8 x+16 $$

Factor completely. Assume that variables in exponents represent positive integers. $$ x^{2}-\left(\frac{1}{x}\right)^{2} $$
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