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Chapter 5: Problem 62

Factor completely. $$ x^{2}(x+3)-4(x+3) $$

Short Answer

Expert verified
The completely factored form is \( (x+3)(x+2)(x-2) \).

Step by step solution

01

Identify Common Factors

Identify the common factor in each term. Notice that both terms contain \(x+3\). Thus, \(x+3\) is a common factor.
02

Factor Out the Common Term

Factor \(x+3\) out of both terms: \[x^{2}(x+3) - 4(x+3) = (x+3)(x^{2} - 4)\]
03

Factor the Difference of Squares

Recognize that \(x^{2} - 4\) is a difference of squares, which can be factored as \((x+2)(x-2)\). Therefore, \[ x^{2} - 4 = (x+2)(x-2) \]
04

Write the Final Factored Form

Substitute the factored form of \(x^{2} - 4\) back into the expression: \[ (x+3)((x+2)(x-2)) \]

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Key Concepts

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

Common Factors
In polynomial expressions, a common factor is an algebraic term that appears in each term of the expression. Identifying and factoring out common factors is the first step in simplification.

Consider the polynomial given in the exercise: \[ x^{2}(x+3)-4(x+3) \]

Here, the term \(x + 3\) appears in both parts of the expression, making it a common factor.

By factoring out \(x + 3\), the expression simplifies to:

\[ x^{2}(x+3) - 4(x+3) = (x+3)(x^{2} - 4) \]

This reveals a new, simpler polynomial that can be further factored, as shown in the next sections.
Difference of Squares
The difference of squares is a special factoring technique used when you have a quadratic expression that subtracts one square term from another square term.

The general form is \[ a^{2} - b^{2} = (a + b)(a - b) \]

In our exercise, the term within the parentheses, \(x^{2} - 4\), fits the pattern of a difference of squares. This is because \[ x^{2} = (x)^{2} \] and \[ 4 = (2)^{2} \]

Applying the difference of squares formula results in:

\[ x^{2} - 4 = (x + 2)(x - 2) \]

Thus, the expression \[ (x + 3)(x^{2} - 4) \] can now be written as \[ (x + 3)(x + 2)(x - 2) \]
Polynomial Factorization
Polynomial factorization involves breaking down a polynomial into a product of simpler polynomials.

The goal is to express the original polynomial as a product of its factors, making it easier to solve or simplify complex expressions.

In our example, we began with:

\[ x^{2}(x + 3) - 4(x + 3) \]

We identified the common factor, which simplified our expression to:

\[ (x + 3)(x^{2} - 4) \]

Then, noticing that \[ x^{2} - 4 \] is a difference of squares, we factored it as \[ (x + 2)(x - 2) \]

Hence, the fully factored form is:

\[ (x + 3)(x + 2)(x - 2) \]

This step-by-step approach allows us to simplify and solve polynomial equations efficiently.

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Most popular questions from this chapter

A Pythagorean triple is a set of three numbers that satisfy the Pythagorean equation. They can be generated by choosing natural numbers \(n\) and \(m\) \(n>m,\) and forming the following three numbers: \(n^{2}+m^{2}, n^{2}-m^{2},\) and \(2 m n .\) Show that these three expressions satisfy the Pythagorean equation.

Solve. Fireworks Displays. Fireworks are typically launched from a mortar with an upward velocity (initial speed) of about \(64 \mathrm{ft} / \mathrm{sec} .\) The height \(h(t)\) in feet, of a "weeping willow" display, \(t\) seconds after having been launched from an 80 -ft high rooftop, is given by $$h(t)=-16 t^{2}+64 t+80 $$ How long will it take the cardboard shell from the fireworks to reach the ground? (image cannot copy) (graph cannot copy)

Perform the indicated operations. $$ \left(8 r^{2}-6 r\right)-(2 r-6)+\left(5 r^{2}-7\right) $$

Let \(f(x)=3 x+1\) and \(g(x)=x^{2}-2 .\) Find the following. The domain of \(f\)

Factor completely. $$ (m-1)^{3}-(m+1)^{3} $$
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