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Chapter 1: Problem 97

Factor the expression completely. $$(x-1)(x+2)^{2}-(x-1)^{2}(x+2)$$

Short Answer

Expert verified
The expression factors to \(3(x-1)(x+2)\).

Step by step solution

01

Identify the Common Factor

The given expression is \((x-1)(x+2)^{2}-(x-1)^{2}(x+2)\). Notice that both terms contain the common factor \((x-1)(x+2)\). We can factor this out from the expression.
02

Factor Out the Common Factor

Factor \((x-1)(x+2)\) out from the expression. This gives us: \[(x-1)(x+2)\big((x+2) - (x-1)\big) \]
03

Simplify the Remaining Expression

Simplify the expression inside the parenthesis: \[(x+2) - (x-1) = x + 2 - x + 1 = 3\] This leaves us with: \[(x-1)(x+2)(3)\]
04

Rearrange the Expression to Show Factoring

The expression can be neatly written as: \[3(x-1)(x+2)\] This is the completely factored form of the original expression.

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Key Concepts

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

Common Factors
In the world of algebra, common factors are elements that appear in all parts of an expression. Recognizing these can simplify a problem considerably. In the given exercise, both terms in the expression \((x-1)(x+2)^{2}-(x-1)^{2}(x+2)\) share a common factor. By identifying the shared component, you can pull it out, significantly shrinking the expression.
  • Check for common elements present in all terms.
  • In this exercise, the common factor is \((x-1)(x+2)\).
After discovery, factor out these elements. For example, factoring out \((x-1)(x+2)\) from the expression makes further simplification possible. By focusing on shared components, you streamline equations, leading to faster and more intuitive problem-solving.
Binomial Multiplication
Understanding binomial multiplication is key to successfully navigating algebra. A binomial is an algebraic expression containing two distinct terms, like \((x-1)\) or \((x+2)\). When multiplying binomials, each term in the first binomial is multiplied by each term in the second.Let's consider the exercise:
  • The expression involves the multiplication of two binomials: \((x-1)\) and \((x+2)\), and variations of these raised to powers, creating products like \((x+2)^{2}\) and \((x-1)^{2}(x+2)\).
  • Notice how these binomials multiply together to form larger polynomials.
Being comfortable with the distributive property (i.e., to expand \((a+b)(c+d)\) into \(ac + ad + bc + bd\)) helps manage binomial multiplication. This groundwork ensures you're equipped to simplify complex equations.
Polynomial Simplification
Once you've handled common factors and binomial multiplication, the next step is simplifying the polynomial. Simplification streamlines expressions into their most manageable form. In our specific exercise, this involves breaking down complex polynomial terms.
  • After factoring out the common elements, what's inside the parentheses becomes simple: \((x+2) - (x-1) = 3\).
  • With the tougher bits out of the way, what's left is rearranging the equation: \((x-1)(x+2)(3)\).
  • This results in the neat and complete factored form: \[3(x-1)(x+2)\].
Simplification is about making an expression as tidy as possible, ensuring it's easy to interpret and utilize. By practicing these principles, algebra becomes a collection of puzzles to solve with confidence and flair.

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Most popular questions from this chapter

Factor the expression completely. (This type of expression arises in calculus when using the "product rule.") $$3(2 x-1)^{2}(2)(x+3)^{1 / 2}+(2 x-1)^{3}\left(\frac{1}{2}\right)(x+3)^{-1 / 2}$$

The left-hand column in the table lists some common algebraic errors. In each case, give an example using numbers that show that the formula is not valid. An example of this type, which shows that a statement is false, is called a counterexample.$$\begin{array}{|c|c|} \hline \text { Algebraic error } & \text { Counterexample } \\ \hline \frac{1}{a}+\frac{1}{b}=\frac{1}{\sqrt{a}+b} & \frac{1}{2}+\frac{1}{2} \neq \frac{1}{2+2} \\ (a+b)^{2}=a^{2}+b^{2} & \\ \sqrt{a^{2}+b^{2}}=a+b & \\ \frac{a+b}{a}=b \\ \left(a^{3}+b^{3}\right)^{1 / 3}=a+b & \\ a^{m} / a^{n}=a^{m / n} & \\ a^{-1 / n}=\frac{1}{a^{n}} & \\ \hline \end{array}$$

Factor the expression completely. $$y^{4}(y+2)^{3}+y^{5}(y+2)^{4}$$

Use a Special Factoring Formula to factor the expression. $$8 s^{3}-125 t^{6}$$

Factor the trinomial. $$(3 x+2)^{2}+8(3 x+2)+12$$
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