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Chapter 3: Problem 35

For Problems \(25-50\), factor completely. $$ 18 a^{2} b+27 a b^{2} $$

Short Answer

Expert verified
The expression is factored as \(9ab(2a + 3b)\).

Step by step solution

01

Identify Common Factors

First, identify the greatest common factor (GCF) of the terms in the expression. Both terms, \(18a^2b\) and \(27ab^2\), have the common factors of \(9\), \(a\), and \(b\). Thus, the GCF is \(9ab\).
02

Factor Out the Greatest Common Factor

Divide each term by \(9ab\) and factor it out of the expression:\[18a^2b + 27ab^2 = 9ab(2a + 3b)\]
03

Verify the Factorization

Multiply \(9ab\) by the terms inside the parenthesis to ensure the expression expands back to the original:\[9ab(2a) = 18a^2b, \quad 9ab(3b) = 27ab^2\]Verification confirms that the factorization is correct.

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Key Concepts

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

Greatest Common Factor
Finding the greatest common factor (GCF) is a key step in factoring polynomials. The GCF is the largest factor that divides each term of the polynomial without a remainder. To find the GCF:
  • List out the factors of each component in the terms.
  • Identify the common factors across all terms.
  • Choose the highest degree of each matching factor.
For example, in the expression \(18a^2b + 27ab^2\), both terms can be divided by \(9\), \(a\), and \(b\). The highest degree for these common factors is \(9ab\). This is because:
  • \(9\) is a factor of \(18\) and \(27\).
  • Each term contains at least one \(a\) and \(b\).
By factoring out the GCF, the expression simplifies considerably, paving the way for further simplification if possible.
Algebraic Expressions
Algebraic expressions are mathematical phrases involving numbers, variables, and operation symbols. They can represent real-world scenarios or more abstract mathematical ideas. In any algebraic expression, understanding the components is crucial:
  • Coefficients: These are the numerical part of the terms. In \(18a^2b\), \(18\) is the coefficient.
  • Variables: Symbols that stand for numbers. Here, \(a\) and \(b\) are variables.
  • Exponents: Indicate how many times a variable is used as a factor. For example, \(a^2\) means \(a\times a\).
Decomposing these expressions by factoring helps find commonalities, like the GCF, making them easier to work with. When factoring, we effectively "break down" the expression into simpler, multiplied parts which can be managed or simplified further.
Mathematical Verification
Mathematical verification is an essential step to confirm that the factorization of an expression is correct. Once an expression has been factored, you must ensure that multiplying these factors returns to the original expression. This step is fundamental for:
  • Ensuring the accuracy of solutions.
  • Building confidence in problem-solving skills.
In the given exercise, we factored \(18a^2b + 27ab^2\) as \(9ab(2a + 3b)\). To verify:
  • Multiply \(9ab\) by \(2a\) to get \(18a^2b\).
  • Multiply \(9ab\) by \(3b\) to get \(27ab^2\).
Both terms return us to the initial algebraic expression, confirming the factorization was accurate. Verification ensures that no errors were made during calculation, and it is a good practice to adopt in all mathematical tasks.

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Most popular questions from this chapter

For Problems 104-109, factor each trinomial and assume that all variables that appear as exponents represent positive integers. $$ 20 x^{2 n}+21 x^{n}-5 $$

Problems \(63-100\) should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using integers. (Objective 4) $$ 2 x y+6 x+y+3 $$

Problems \(63-100\) should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using integers. (Objective 4) $$ x^{3}+125 $$

For Problems \(1-54\), solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. (Objective 1) $$ w^{2}-4 w=5 $$

Consider the following two solutions for the equation \((x+3)(x-4)=(x+3)(2 x-1)\) Solution A $$ \begin{aligned} &(x+3)(x-4)=(x+3)(2 x-1) \\ &(x+3)(x-4)-(x+3)(2 x-1)=0 \\ &(x+3)[x-4-(2 x-1)]=0 \\ &(x+3)(x-4-2 x+1)=0 \\ &(x+3)(-x-3)=0 \end{aligned} $$ $$ \begin{aligned} x+3 &=0 & & \text { or } & &-x-3=0 \\ x &=-3 & & \text { or } & &-x=3 \\ x &=-3 & & \text { or } & & x=-3 \end{aligned} $$ The solution set is \(\\{-3\\}\). Solution B $$ \begin{aligned} (x+3)(x-4) &=(x+3)(2 x-1) \\ x^{2}-x-12 &=2 x^{2}+5 x-3 \\ 0 &=x^{2}+6 x+9 \\ 0 &=(x+3)^{2} \\ x+3 &=0 \\ x &=-3 \end{aligned} $$ The solution set is \(\\{-3\\}\). Are both approaches correct? Which approach would you use, and why?
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