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Chapter 10: Problem 87

Factor: \(y^{2}-6 y+9-25 x^{2}\) (Section 6.5, Example 8)

Short Answer

Expert verified
The factored form of the expression \(y^{2}-6 y+9-25 x^{2}\) is \((y-3+5x)(y-3-5x)\).

Step by step solution

01

Identify the quadratics

Looking at the expression \(y^{2}-6 y+9-25 x^{2}\), you can identify that \((y^{2}-6y+9)\) and \((-25x^{2})\) are the two quadratics inside the expression. The first quadratic is a perfect square trinomial, and the second quadratic is a perfect square.
02

Factor the Perfect Square Trinomial

The perfect square trinomial \(y^{2}-6y+9\) can be factored as \((y-3)^{2}\), and the perfect square \(-25x^{2}\) can be factored as \(-5x)^{2}\).
03

Apply the difference of squares rule

Using the difference of squares rule \(a^{2}-b^{2} = (a+b)(a-b)\), the factored form of the expression is \((y-3+5x)(y-3-5x)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Perfect Square Trinomial
Understanding a perfect square trinomial is crucial when approaching algebraic factoring. A perfect square trinomial is an algebraic expression of the form \(a^{2} + 2ab + b^{2}\) or \(a^{2} - 2ab + b^{2}\), where \(a\) and \(b\) are real numbers.

When factored, it simplifies to \( (a + b)^{2}\) or \( (a - b)^{2}\) respectively. To identify this pattern:
  • Look for a trinomial where the first and last terms are perfect squares.
  • Confirm that the middle term is twice the product of the square roots of the first and last terms.
For example, in the expression \(y^{2}-6y+9\), \(y^{2}\) is the square of \(y\), and \(9\) is the square of \(3\). The middle term, \(6y\), is twice the product of \(y\) and \(3\). Therefore, it factors to \( (y-3)^{2}\), demonstrating it's a perfect square trinomial.
Difference of Squares
The difference of squares is another fundamental concept in algebra, which refers to subtracting one perfect square from another. The rule for factoring a difference of squares is straightforward: an expression \(a^{2} - b^{2}\) factors into \( (a + b)(a - b)\).

To apply this rule:
  • Identify each term as a perfect square.
  • Take the square root of each term to determine \(a\) and \(b\).
  • Write the expression as the product of two binomials.
Building on the previous section, after factoring the perfect square trinomial in \(y^{2}-6y+9-25x^{2}\), we are left with \( (y-3)^{2}- (5x)^{2}\), which highlights a difference of squares situation. This factors further to \( (y-3+5x)(y-3-5x)\), which demonstrates the difference of squares at work.
Algebraic Expressions
An algebraic expression is a combination of numbers, variables (like \(x\) or \(y\)), and arithmetic operations (such as \(+\), \( -\), \(\times\), and \(\div\)). Algebraic expressions can range from simple, like \(2x + 3\), to more complex forms, such as the one in our example, \(y^{2}-6y+9-25x^{2}\).

To manipulate these expressions effectively, one must understand:
  • The operations and how they can be applied or reversed.
  • Terms, factors, and coefficients within the expressions.
  • Patterns of common factoring techniques such as factoring out a greatest common factor, factoring quadratics, and the difference of squares.
Mastering these concepts allows us to transform complex algebraic expressions into simplified forms, often aiding in solving equations or understanding relationships in a given problem.

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Most popular questions from this chapter

What are conjugates? Give an example with your explanation.

In Exercises \(105-112,\) add or subtract as indicated. Begin by rationalizing denominators for all terms in which denominators contain radicals. $$\frac{5}{\sqrt{2}+\sqrt{7}}-2 \sqrt{32}+\sqrt{28}$$

In Exercises \(105-112,\) add or subtract as indicated. Begin by rationalizing denominators for all terms in which denominators contain radicals. $$\sqrt{2}+\frac{1}{\sqrt{2}}$$

Exercises \(88-90\) will help you prepare for the material covered in the next section. Simplify: \((-5+7 x)-(-11-6 x)\)

Use a graphing utility to solve each radical equation. Graph each side of the equation in the given viewing rectangle. The equation's solution set is given by the \(x\) -coordinate(s) of the point (s) of intersection. Check by substitution. $$\begin{aligned} &\sqrt{x}+4=2\\\ &[-2,18,1] \text { by }[0,10,1] \end{aligned}$$
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