Problem 63
Question
Factor each trinomial completely. Some of these trinomials contain a greatest common factor (other than 1). Don't forget to factor out the GCF first. $$ 162-45 m+3 m^{2} $$
Step-by-Step Solution
Verified Answer
The factored form is \(3(m - 9)(m - 6)\).
1Step 1: Identify the GCF
First, examine the coefficients 162, -45, and 3 to determine their greatest common factor (GCF). The GCF of these numbers is 3.
2Step 2: Factor out the GCF
Factor out the GCF from the trinomial:\[ 3(54 - 15m + m^2) \]
3Step 3: Rearrange the Terms
Rearrange the terms inside the parenthesis in descending order of the exponent on \(m\):\[ 3(m^2 - 15m + 54) \]
4Step 4: Factor the Quadratic
Now, factor the quadratic expression. Look for two numbers that multiply to 54 and sum to -15. The numbers are -9 and -6. Thus, the factors are plain parentheses: \( (m - 9)(m - 6) \).
5Step 5: Write the Factored Form
Combine all parts to write the completely factored form:\[ 3(m - 9)(m - 6) \]
Key Concepts
greatest common factorquadratic expressionsfactoring techniques
greatest common factor
The greatest common factor (GCF) is the largest number that divides two or more numbers without leaving a remainder. Identifying the GCF is a critical first step when factoring, because it simplifies the problem.
To find the GCF of 162, -45, and 3, examine the factors of each number:
- 162 has factors 1, 2, 3, 6, 9, 18, 27, 54, 81, 162.
- -45 has factors 1, 3, 5, 9, 15, 45 (note that we deal with positive factors only when finding the GCF).
- 3 is a factor only of 3.
quadratic expressions
Quadratic expressions are polynomials of degree two, typically written in the form \( ax^2 + bx + c \). These expressions are pivotal in algebra and appear in a variety of problems.Our original trinomial, after factoring out the GCF, is \( m^2 - 15m + 54 \). Here, \( m^2 \) is the quadratic term, \(-15m\) is the linear term, and 54 is the constant term.Quadratic expressions can often be factored into the product of two binomials. This involves finding two numbers that multiply to give the constant term and add to give the coefficient of the linear term. Recognizing this structure is essential as it allows you to break down complex problems into more manageable pieces. Understanding the anatomy of a quadratic expression is key to applying factoring techniques effectively.
factoring techniques
Factoring techniques are strategies used to break down polynomials into simpler, multiplicative components. For quadratic expressions, a common technique is identifying two numbers that multiply to the last term and add to the middle coefficient.In our example, the quadratic expression is \( m^2 - 15m + 54 \). We look for two numbers that multiply to 54 (the constant) and add up to -15 (the coefficient of \( m \)). The numbers that work here are -9 and -6.Thus, the quadratic expression can be factored as \((m - 9)(m - 6)\). This technique reduces complex expressions into products of simpler expressions, which can then be further simplified or analyzed. Factoring methods like trial and error, grouping, or the use of special products (such as squares of binomials or differences of squares) are valuable tools in solving polynomial equations.
Other exercises in this chapter
Problem 63
Factor. $$ x^{3} y^{2}-27 y^{2} $$
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Factor each four-term polynomial by grouping. If this is not possible, write "not factorable by grouping." $$ 6 x^{3}-4 x^{2}+15 x-10 $$
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Factor each trinomial completely. See Examples 1 through 7. \(6 y^{3}-8 y^{2}-30 y\)
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Solve each equation. $$ 7 x^{3}-7 x=0 $$
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