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Chapter 6: Problem 111

Factor each trinomial completely. \(4 x^{2}(y-1)^{2}+10 x(y-1)^{2}+25(y-1)^{2}\)

Short Answer

Expert verified
Factored form is \((y-1)^{2}(2x+5)^{2}\).

Step by step solution

01

Identify the Common Factor

The given expression is: \[4x^{2}(y-1)^{2} + 10x(y-1)^{2} + 25(y-1)^{2}\]Notice that each term contains \((y-1)^{2}\). So, \((y-1)^{2}\) is a common factor. Factor \((y-1)^{2}\) out of the expression:\[(y-1)^{2}(4x^{2} + 10x + 25)\]
02

Factor the Quadratic Expression

Focus on factoring the quadratic expression inside the parentheses: \[4x^{2} + 10x + 25\]Look for two numbers that multiply to \(4 \times 25 = 100\) and add to 10. These numbers are 5 and 5. So, use the format of \((Ax + B)^2\):\[(2x + 5)^2\]
03

Combine the Factors

Now substitute back the factorization into the original expression:\[(y-1)^{2}(2x + 5)^{2}\]This is the fully factored form of the original trinomial.

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Key Concepts

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

Understanding Common Factors
A common factor is something that appears in each term of an expression and can be factored out to simplify the expression. In the context of polynomials and trinomials, identifying a common factor is a crucial first step. In the given problem, the original expression is \[4x^{2}(y-1)^{2} + 10x(y-1)^{2} + 25(y-1)^{2}\]Here, \((y-1)^{2}\)appears in each term, acting as a common factor. By factoring it out, you transform the expression into\[(y-1)^{2}(4x^{2} + 10x + 25)\].Finding common factors not only simplifies the solving process but also makes it easier to manage more complex expressions. To spot common factors:
  • Look for variables or numbers present in all terms.
  • Factor them out to reduce the complexity.
Quadratic Expressions
Quadratic expressions are polynomial expressions of degree two. They take the general form \(ax^2 + bx + c\),where \(a\), \(b\), and \(c\) are constants. These are the building blocks of the expression we aim to factor.In the given problem, once the common \((y-1)^{2}\)was factored out, we were left with a quadratic expression: \[4x^{2} + 10x + 25\].Quadratics may be challenging at first due to their non-linear nature, but recognizing patterns can simplify factoring. For simplicity, it can help to:
  • Assess whether the expression can be factored further into simpler binomial products.
  • Search for pairs of numbers that fulfill necessary multiplication and addition properties, connecting to terms in the expression.
Understanding how quadratic expressions are structured is key to mastering various factoring techniques.
Exploring Factoring Techniques
There are several factoring techniques suited to different types of polynomials. For trinomials like \(4x^{2} + 10x + 25\), the approach often involves:
  • Identifying patterns for perfect squares or looking for factors of the constant term that satisfy both the middle term's coefficient and multiplication requirements.
  • Using the format \((Ax + B)^2\).
In this case, we identified numbers \(5 \times 5 = 25\)and \(5 + 5 = 10\)which led us to \((2x + 5)\)^2.This is an efficient technique that follows analysis of symmetry and pattern recognition. By factoring using these methods, expressions are simplified into products of binomials, making them easier to work with either in further mathematical operations or analysis. Effectively applying these techniques allows one to handle trinomials with greater ease and success.

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Most popular questions from this chapter

Factor each trinomial completely. See Examples I through II and Section 6.2. \(25 x^{2}-60 x y+36 y^{2}\)

Factor each trinomial completely. $$ x^{2}+\frac{1}{2} x+\frac{1}{16} $$

Multiply the following. See Section 5.4 \((z-2)\left(z^{2}+2 z+4\right)\)

Factor each binomial completely. \(8 m^{3}+64\)

Solve each equation. See Examples I and 2. $$ (3 x-2)(5 x+1)=0 $$
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