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Chapter 6: Problem 45

Factor each trinomial completely. See Examples 1 through 7. \(6 x^{2} y^{2}-2 x y^{2}-60 y^{2}\)

Short Answer

Expert verified
\( 2y^2(x - 10)(3x + 9) \)

Step by step solution

01

Identify the Common Factor

Look at each term in the trinomial: - First term: \( 6x^2y^2 \)- Second term: \( -2xy^2 \)- Third term: \( -60y^2 \)Notice that each term contains \( y^2 \) as a common factor. Factor \( y^2 \) out of the entire expression.
02

Factor Out the Common Term

Factor \( y^2 \) from each term:\[ 6x^2y^2 - 2xy^2 - 60y^2 = y^2 (6x^2 - 2x - 60) \]
03

Factor the Remaining Trinomial

Look at the trinomial \( 6x^2 - 2x - 60 \) and determine if it can be factored further. First, observe that there is a common factor of 2:\[ 6x^2 - 2x - 60 = 2(3x^2 - x - 30) \]
04

Factor Completely

Next, factor the quadratic trinomial \( 3x^2 - x - 30 \). Find two numbers that multiply to \( 3 \cdot (-30) = -90 \) and add to \(-1\). These numbers are \( -10 \) and \( 9 \). Rewrite \(-x\) as \(-10x + 9x\) and factor by grouping:\[ 3x^2 - x - 30 = 3x^2 - 10x + 9x - 30 \]\[ = (3x^2 - 10x) + (9x - 30) \]Now factor by grouping:\[ = 1x(3x - 10) + 3(3x - 10) \]\[ = (3x + 3)(x - 10) \]This factors to \((3x + 9)(x - 10)\). Further simplify:\[ 2(3x^2 - x - 30) = 2(x - 10)(3x + 9) \]
05

Final Expression

Now substitute back to include all the factored terms:\[ 6x^2y^2 - 2xy^2 - 60y^2 = y^2 \cdot 2 \cdot (x - 10)(3x + 9) \]Simplify the constant terms:\[ = 2y^2(x - 10)(3x + 9) \]

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Key Concepts

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

Common Factor
A common factor is a number or expression that divides two or more terms perfectly without leaving a remainder. In the context of factoring trinomials, identifying a common factor is usually the first step. For example, consider the trinomial \(6x^2y^2 - 2xy^2 - 60y^2\):
  • The common factor here is \(y^2\). This term is present in all parts of the trinomial and can be pulled out of the expression.
  • Once you identify the common factor, factor it out to make the remaining expression simpler.
By factoring out \(y^2\), you transform the original trinomial into \(y^2 (6x^2 - 2x - 60)\). This sets the stage for further factoring.
Quadratic Trinomial
A quadratic trinomial is an expression composed of three terms, with the highest degree term being quadratic. The standard form of a quadratic trinomial is \(ax^2 + bx + c\). In our example, the expression \(6x^2 - 2x - 60\) is a quadratic trinomial.
  • To factor a quadratic trinomial, start by identifying patterns or use systematic methods like factoring by grouping or trial and error.
  • Quadratic trinomials can often be factored into two binomials, though some may be prime and not factorable by normal means.
In this case, further factoring was required, resulting in the expression \(2(3x^2 - x - 30)\). The goal is to break down the trinomial into smaller, manageable components.
Factoring by Grouping
Factoring by grouping is a method used to simplify and ultimately factor complex polynomials. This method works well when an expression can be split into smaller groups that have their own common factors. Let's take a closer look at the expression \(3x^2 - x - 30\).
  • Start by rewriting \(-x\) as two terms that add to \(-x\) and multiply to the product of the leading coefficient and the constant term (here, it鈥檚 \(-90\)).
  • In our example, these numbers are \(-10\) and \(9\), which results in rephrasing the expression as \(3x^2 - 10x + 9x - 30\).
To continue, group the terms: \((3x^2 - 10x) + (9x - 30)\). Factoring the common factor from each group gives us: \(x(3x - 10) + 3(3x - 10)\), and finally, we can factor the expression as \((x - 10)(3x + 9)\). This method ensures each group is simple to factor, leading to the complete factorization.

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Most popular questions from this chapter

Factor. $$ z^{3}-1 $$

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