Problem 27
Question
For the following problems, the first quantity represents the product and the second quantity a factor. Find the other factor. $$ 6 b^{2}-6 b-3,3 $$
Step-by-Step Solution
Verified Answer
Answer: The other factor is $(2b^2 - 2b - 1)$.
1Step 1: Write down the given product and factor
We are given the product:
$$
6b^2 - 6b - 3
$$
and the factor:
$$
3
$$
2Step 2: Divide the product by the known factor
Divide the given product \((6b^2 - 6b - 3)\) by the known factor \((3)\):
$$
\frac{6b^2 - 6b - 3}{3}
$$
3Step 3: Simplify the result
Simplify the quotient to find the other factor:
$$
\frac{6b^2 - 6b - 3}{3} = 2b^2 - 2b - 1
$$
The other factor is:
$$
2b^2 - 2b - 1
$$
Key Concepts
Polynomial DivisionSimplifying ExpressionsAlgebraic Factors
Polynomial Division
Polynomial division is a process similar to long division in arithmetic, but it involves dividing polynomials instead of numbers. It's used to simplify expressions or to factor polynomials, and one of its applications is to find the remainder of a division or to express a polynomial as a product of simpler factors.
In the exercise example, the polynomial \(6b^2 - 6b - 3\) is divided by the monomial \(3\). This is a simpler case of polynomial division where each term in the polynomial is divided by a single term. You essentially apply the distributive property in reverse to factor out the \(3\). Since each term of the polynomial is divisible by \(3\), we are able to express the polynomial as the product of \(3\) and another polynomial.
In the exercise example, the polynomial \(6b^2 - 6b - 3\) is divided by the monomial \(3\). This is a simpler case of polynomial division where each term in the polynomial is divided by a single term. You essentially apply the distributive property in reverse to factor out the \(3\). Since each term of the polynomial is divisible by \(3\), we are able to express the polynomial as the product of \(3\) and another polynomial.
Simplifying Expressions
Simplifying expressions in algebra involves altering the form of an expression without changing its value. The goal of simplification is to make the expression easier to understand and work with. It often involves combining like terms, factoring, canceling common factors, or expanding products.
In our exercise, the simplification process involves dividing each term of \(6b^2 - 6b - 3\) by \(3\). This division results in a more straightforward expression, \(2b^2 - 2b - 1\), which is the simplified form. Simplifying expressions can make it easier to identify properties such as the roots of a polynomial or potential symmetries and can also be a stepping stone to further operations, like derivatives in calculus.
In our exercise, the simplification process involves dividing each term of \(6b^2 - 6b - 3\) by \(3\). This division results in a more straightforward expression, \(2b^2 - 2b - 1\), which is the simplified form. Simplifying expressions can make it easier to identify properties such as the roots of a polynomial or potential symmetries and can also be a stepping stone to further operations, like derivatives in calculus.
Algebraic Factors
In the algebraic context, factors are expressions that can be multiplied together to get another expression, the product. Finding algebraic factors is an important skill, as it allows us to break down complex polynomials into simpler, more manageable pieces. These factors can be numbers, variables, or more complicated expressions.
The exercise dealt with factoring by division, where the given product \(6b^2 - 6b - 3\) is divided by one of its factors, \(3\), to find the other factor, \(2b^2 - 2b - 1\). Factoring helps in solving polynomial equations, finding zeros of polynomial functions, and in simplification of rational expressions. It's one of the foundational techniques used in higher-level mathematics such as calculus, where factoring enables integration by partial fractions or finding limits using algebraic manipulation.
The exercise dealt with factoring by division, where the given product \(6b^2 - 6b - 3\) is divided by one of its factors, \(3\), to find the other factor, \(2b^2 - 2b - 1\). Factoring helps in solving polynomial equations, finding zeros of polynomial functions, and in simplification of rational expressions. It's one of the foundational techniques used in higher-level mathematics such as calculus, where factoring enables integration by partial fractions or finding limits using algebraic manipulation.
Other exercises in this chapter
Problem 27
For the following problems, factor the trinomials when possible. $$ a^{2}-12 a+20 $$
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For the following problems, use the grouping method to factor the polynomials. Some polynomials may not. be factorable using the grouping method. $$ 5 x^{2} y^{
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In the following problems, the first quantity represents the product and the second quantity represents a factor of that product. Find the other factor. $$ 12 a
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For the following problems, factor the binomials. $$ 121 a^{2}-9 $$
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