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

Ch. R - Review of Basic Concepts

Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook

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Lial 13th Edition

Ch. R - Review of Basic Concepts

Problem 112

Chapter 1, Problem 112

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

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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.

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.

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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