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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 40

Find each product. (3y-5)(3y+5)(9y2-25)

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1
Recognize that the expression is a product of three factors: \( (3y - 5)(3y + 5)(9y^2 - 25) \). Notice that the first two factors form a difference of squares pattern.
Apply the difference of squares formula to the first two factors: \( (a - b)(a + b) = a^2 - b^2 \). Here, \(a = 3y\) and \(b = 5\), so \( (3y - 5)(3y + 5) = (3y)^2 - 5^2 = 9y^2 - 25 \).
Substitute the result back into the expression, so now you have \( (9y^2 - 25)(9y^2 - 25) \), which is the square of \( (9y^2 - 25) \).
Rewrite the expression as \( (9y^2 - 25)^2 \) and recognize this as a perfect square binomial.
Expand \( (9y^2 - 25)^2 \) using the formula \( (A - B)^2 = A^2 - 2AB + B^2 \), where \( A = 9y^2 \) and \( B = 25 \). Write out the expanded form without simplifying the coefficients.

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Key Concepts

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

Polynomial Multiplication

Polynomial multiplication involves multiplying each term in one polynomial by every term in the other polynomial(s) and then combining like terms. This process extends the distributive property and is essential for expanding expressions like (3y - 5)(3y + 5).
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Difference of Squares

The difference of squares is a special product formula: (a - b)(a + b) = a² - b². Recognizing this pattern simplifies multiplication, as seen in (3y - 5)(3y + 5), which equals 9y² - 25, reducing the complexity of the problem.
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Combining Like Terms

After multiplying polynomials, combining like terms means adding or subtracting terms with the same variable and exponent. This step simplifies the expression into its standard form, making it easier to interpret or use in further calculations.
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