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Chapter 6: Problem 90

Factor by grouping. \(b^{3}-2+5 a b^{3}-10 a\)

Short Answer

Expert verified
\( (1 + 5a)(b^{3} - 2) \)

Step by step solution

01

Group the terms

Group the given polynomial into pairs that can be factored individually. The polynomial is given as: \( b^{3} - 2 + 5a b^{3} - 10a \). The pairs can be grouped as follows: \( (b^{3} + 5a b^{3}) + (-2 - 10a) \).
02

Factor out common terms in each group

Factor out the common terms from each group. For the first group \( ( b^{3} + 5a b^{3} ) \), factor out \( b^{3} \). For the second group \( (-2 - 10a) \), factor out \( -2 \). This results in: \( b^{3}(1 + 5a) - 2(1 + 5a) \).
03

Factor out the common binomial

Notice that \( (1 + 5a) \) is a common binomial in both terms. Factor it out: \( (1 + 5a)(b^{3} - 2) \).

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Key Concepts

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

Polynomial Factorization
Polynomial factorization is the process of breaking down a polynomial into a product of simpler polynomials. Think of it as reverse distributing. We work to simplify the polynomial into factors that can be multiplied to obtain the original polynomial. For example, in the exercise given, we start with the polynomial: \( b^{3} - 2 + 5a \ b^{3} - 10a \). By grouping like terms and factoring common factors out, we simplify the polynomial as shown in the solution steps.
Common Binomial
A common binomial is a term that appears in multiple parts of the polynomial. Recognizing it helps in further simplifying the expression. In this exercise, after grouping and factoring out common terms, we notice that \( (1 + 5a) \) appears in both groups. This common binomial can then be factored out, leading us to the fully factored form: \( (1 + 5a)(b^{3} - 2) \). Once you find a common binomial in separate groups, it means we can pull it out just like we would with a common monomial term.
Algebraic Expressions
Algebraic expressions contain numbers, variables, and operations. Understanding how to manipulate these expressions鈥攊ncluding factorization by grouping鈥攊s a crucial algebra skill. In the exercise, we start with the algebraic expression \( b^{3} - 2 + 5a \ b^{3} - 10a \). By rearranging and grouping terms that share factors, we systematically simplify the expression. Breaking down complex algebraic expressions in this way can make them more manageable and reveal simplifications that aren't immediately obvious.

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Most popular questions from this chapter

Galileo 's formula describing the motion of freely falling objects is $$ d=16 t^{2} $$ The distance d in feet an object falls depends on the time \(t\) elapsed, in seconds. (This is an example of an important mathematical concept, the function.) When you substituted 256 for \(d\) and solved the formula for \(t\) in Exercise \(79,\) you should have found two solutions: 4 and \(-4 .\) Why doesn't \(-4\) make sense as an answer?

Factor completely. $$ 5 m^{5}+25 m^{4}-40 m^{2} $$

Solve each equation, and check your solutions. $$ (4 y-3)^{3}-9(4 y-3)=0 $$

Factor each trinomial completely. $$ 4 x^{4}+2 x^{3}+x^{2} $$

Find each product. \(2 x^{2}\left(x^{2}+3 x+5\right)\)
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