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Ch. R - Review of Basic Concepts
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 1, Problem 112

Factor by any method. See Examples 1–7. (3a+5)2-18(3a+5)+81

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1
Recognize that the expression is a quadratic in terms of the binomial \(3a + 5\). Let \(x = 3a + 5\) to simplify the expression to \(x^2 - 18x + 81\).
Rewrite the expression using the substitution: \(x^2 - 18x + 81\). Now, focus on factoring this quadratic expression.
Look for two numbers that multiply to \(81\) and add up to \(-18\). These numbers will help factor the quadratic into the form \((x - m)(x - n)\).
Once the quadratic is factored as \((x - m)(x - n)\), substitute back \(x = 3a + 5\) to get the factors in terms of \(a\).
Write the final factored form as \((3a + 5 - m)(3a + 5 - n)\), which is the factorization of the original expression.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Substitution Method

The substitution method involves replacing a complex expression with a single variable to simplify the factoring process. In this problem, letting x = (3a + 5) transforms the expression into a quadratic form, making it easier to factor.
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Factoring Quadratic Expressions

Factoring quadratics means rewriting a quadratic expression as a product of two binomials. Recognizing the standard form ax^2 + bx + c allows you to find factors of c that add up to b, or use methods like completing the square or the quadratic formula.
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Difference of Squares and Perfect Square Trinomials

Understanding special factoring patterns like difference of squares and perfect square trinomials helps quickly factor expressions. For example, recognizing if the quadratic is a perfect square trinomial can simplify factoring without trial and error.
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