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17–24 ? Use a Factoring Formula to factor the expression. $$ 9 a^{2}-16 $$

Short Answer

Expert verified
The expression factors to \((3a - 4)(3a + 4)\).

Step by step solution

01

Identify the Formula to Use

The expression given is a difference of squares: \(9a^2 - 16\). Recall the formula for factoring a difference of squares: \(x^2 - y^2 = (x - y)(x + y)\).
02

Express as a Difference of Squares

Identify \(x\) and \(y\) such that \(x^2 = 9a^2\) and \(y^2 = 16\). Therefore, \(x = 3a\) and \(y = 4\).
03

Apply the Difference of Squares Formula

Substitute the values into the formula: \(9a^2 - 16 = (3a)^2 - (4)^2 = (3a - 4)(3a + 4)\).
04

Simplify the Expression

The expression \((3a - 4)(3a + 4)\) is already simplified as it is a direct application of the difference of squares formula.

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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 special pattern in algebra, characterized by an expression in the form of \( a^2 - b^2 \). This pattern is called a 'difference' because it involves subtraction. It consists of two terms that are both perfect squares. A perfect square is a number or expression that can be expressed as the product of another number or expression with itself. This pattern stands out because it can be factored easily using the formula: \( (a - b)(a + b) \). This means that the difference of two squares can always be split into two binomials.

For example, in the expression \( 9a^2 - 16 \), both \( 9a^2 \) and \( 16 \) are perfect squares. Specifically, \( 9a^2 \) is \( (3a)^2 \) and \( 16 \) is \( 4^2 \). Therefore, this expression is a difference of squares and can be factored by recognizing each square and applying the formula.
Polynomial Expressions
A polynomial expression is a mathematical expression involving sums and products of variables and coefficients. The variables are often referred to as non-constant placeholders, such as \( x \) or \( a \), and coefficients are their constant multipliers. Polynomial expressions can have different degrees, which are determined by the highest power of the variable present.

In our example, \( 9a^2 - 16 \) is a binomial polynomial of degree 2, even though it appears different because of the subtraction. Typically, polynomial terms with subtraction are rewritten in more manageable forms using identities like the difference of squares, allowing for easier manipulation and factoring.
  • Binomial: A polynomial with two terms.
  • Degree: Determined by the highest exponent in the expression.
  • Coefficients: Numerical factors multiplying the variables.
Factoring Techniques
Factoring is the process of breaking down a complex expression into simpler ones which, when multiplied together, give back the original expression. This is akin to finding the 'building blocks' of the polynomial. There are several methods and techniques used in factoring:

1. **Common Factor:** Look for terms that have common factors and factor them out.
2. **Difference of Squares:** As discussed, this specific technique handles expressions of the form \( a^2 - b^2 \). This is quick and efficient for quadratics missing the linear term due to the distinct pattern they form.
3. **Trinomials:** Often factored by checking for factors of the constant term that add up to the coefficient of the linear term.

In our example of \( 9a^2 - 16 \), recognizing the expression as a difference of squares allows us to factor it into \((3a - 4)(3a + 4)\). This was done by identifying that \( 9a^2 \) is \((3a)^2\) and \( 16 \) is \(4^2\), hence applied the difference of squares technique to simplify the expression in a step-by-step manner.

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Most popular questions from this chapter

\(89-96\) m State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ \frac{2}{4+x}=\frac{1}{2}+\frac{2}{x} $$

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Factoring \(x^{4}+a x^{2}+b\) A trinomial of the form \(x^{4}+a x^{2}+b\) can sometimes be factored easily. For example \(x^{4}+3 x^{2}-4=\left(x^{2}+4\right)\left(x^{2}-1\right) .\) But \(x^{4}+3 x^{2}+4\) cannot be factored in this way. Instead, we can use the following method. \(x^{4}+3 x^{2}+4=\left(x^{4}+4 x^{2}+4\right)-x^{2} \quad\) Add and subtract \(x^{2}\) \(=\left(x^{2}+2\right)^{2}-x^{2} \quad\) Factor perfect square \(=\left[\left(x^{2}+2\right)-x\right]\left[\left(x^{2}+2\right)+x\right] \quad\) Difference of squares \(=\left(x^{2}-x+2\right)\left(x^{2}+x+2\right)\) Factor the following using whichever method is appropriate. (a) \(x^{4}+x^{2}-2\) (b) \(x^{4}+2 x^{2}+9\) (c) \(x^{4}+4 x^{2}+16\) (d) \(x^{4}+2 x^{2}+1\)

\(89-96\) m State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ \frac{b}{b-c}=1-\frac{b}{c} $$

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