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Chapter 6: Problem 193

Factor completely using the difference of squares pattern, if possible. $$ a^{2}+6 a+9-9 b^{2} $$

Short Answer

Expert verified
\((a + 3 - 3b)(a + 3 + 3b)\)

Step by step solution

01

- Identify Perfect Squares

Rewrite the expression to group potential perfect squares: \[ a^2 + 6a + 9 - 9b^2 \]. Notice that \(a^2\) and \(9b^2\) are perfect squares.
02

- Recognize Quadratic Form

Identify and isolate the quadratic trinomial: \[ a^2 + 6a + 9 \]. Recognize this as a perfect square trinomial because \(a^2\) and \(9\) are perfect squares, and \(6a\) fits the middle term as \(2 \times a \times 3\).
03

- Factor the Perfect Square Trinomial

Factor \(a^2 + 6a + 9\) into \((a + 3)^2\). The expression now looks like: \[ (a + 3)^2 - 9b^2 \].
04

- Apply the Difference of Squares Formula

Use the difference of squares pattern \(x^2 - y^2 = (x - y)(x + y)\) to factor \((a + 3)^2 - (3b)^2\). Let \(x = a + 3\) and \(y = 3b\). The expression becomes: \[ ((a + 3) - 3b)((a + 3) + 3b) \].
05

- Simplify the Factored Form

Simplify the final factored form: \[ ((a + 3) - 3b)((a + 3) + 3b) \]. Therefore, the fully factored expression is: \[ (a + 3 - 3b)(a + 3 + 3b) \].

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Key Concepts

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

Difference of Squares
The difference of squares is a fundamental concept in polynomial factorization. It involves two terms that are both perfect squares, separated by a subtraction operation.
When you have an expression like \(x^2 - y^2\), it can be factored into \((x - y)(x + y)\).
This is useful for simplifying expressions and solving equations.
In the provided exercise, the expression \((a + 3)^2 - (3b)^2\) is factored using this method.
Recognize the pattern and apply the formula to factor the polynomial effectively.
Perfect Square Trinomial
A perfect square trinomial is an expression that arises from squaring a binomial.
For instance, the expression \(a^2 + 6a + 9\) comes from squaring \(a + 3\).
  • The first term \(a^2\) is the square of the first term of the binomial.

  • The last term \9\ is the square of the second term of the binomial.

  • The middle term \6a\ is twice the product of the two terms of the binomial, here \(2 \times a \times 3\).
This pattern helps to recognize and factor trinomials easily.
In the given exercise, this recognition allows us to write \(a^2 + 6a + 9\) as \((a + 3)^2\).
Quadratic Form
The quadratic form involves expressions that fit the pattern \(ax^2 + bx + c\).
These are second-degree polynomials with specific properties.
To factor them efficiently, look for patterns like the perfect square trinomial.
In the exercise, the expression \(a^2 + 6a + 9\) fits this form.
Identifying such patterns can simplify the factorization process.
This step-by-step approach reveals the structure of the polynomial and leads us to its factors.
Polynomial Factorization
Polynomial factorization involves breaking down a polynomial into a product of simpler polynomials.
This makes solving equations and simplifying expressions easier.
Different techniques help achieve this:
  • Factoring by grouping

  • Using special patterns like the difference of squares and perfect square trinomials

  • General methods like the quadratic formula
In our exercise, we used these techniques to factor the given polynomial step by step.
First, recognize patterns and rewrite the expression accordingly.
Then apply relevant formulas to achieve the final factored form.

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Most popular questions from this chapter

In the following exercises, factor using the 'ac' method. $$ 5 s^{2}-9 s+4 $$

In the following exercises, factor completely using trial and error. $$ 5 x^{2}-17 x+6 $$

In the following exercises, foctor using substitution. $$ (x-2)^{2}-3(x-2)-54 $$

In the following exercises, factor each trinomial of the form \(x^{2}+b x+c .\) $$ a^{2}+25 a+100 $$

In the following exercises, factor each trinomial of the form \(x^{2}+b x y+c y^{2} .\) $$ x^{2}-2 x y-80 y^{2} $$
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