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Chapter 5: Problem 22

Factor each polynomial. $$ x^{2}+10 x+25 $$

Short Answer

Expert verified
(x+5)^2

Step by step solution

01

Identify the structure of the polynomial

Notice that the polynomial is in the form of a quadratic equation, which can potentially be factored into the form \((x + a)^2\). So, identify if it represents a perfect square trinomial.
02

Recognize perfect square trinomial pattern

The general form of a perfect square trinomial is \(a^2 + 2ab + b^2\). Let's check if our polynomial \(x^2 + 10x + 25\) fits this form.
03

Decompose the middle term and constant

Identify \(a\) and \(b\) so that \(2ab = 10x\) and \(b^2 = 25\). Here, a is equal to \(x\) and b is equal to \(5\) because \(2(1)(5) = 10\) and \(5^2 = 25\).
04

Write as square of binomial

Since the polynomial fits the pattern, write the expression as a squared binomial: \((x+5)^2\).

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Key Concepts

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

Perfect Square Trinomial
A perfect square trinomial is a special type of quadratic polynomial. It has the form \(a^2 + 2ab + b^2\). This structure is very convenient because it can always be factored into the square of a binomial expression.
For example, in the polynomial \(x^2 + 10x + 25\), we recognize that it matches the form \(a^2 + 2ab + b^2\), where \(a = x\) and \(b = 5\).
Therefore, \(x^2 + 10x + 25\) is a perfect square trinomial. It can be factored into \((x + 5)^2\). Understanding this concept helps simplify many algebraic expressions.
Quadratic Equations
Quadratic equations are polynomials of degree two. Their general form is \(ax^2 + bx + c = 0\). These equations are fundamental in algebra and appear in various fields of study, such as physics, engineering, and economics.
Factoring is one way to solve quadratic equations, allowing us to find the values of \(x\) that satisfy the equation. Identifying and working with special cases like perfect square trinomials can make factoring much easier.
Factoring Techniques
Factoring polynomials involves breaking them down into simpler terms, or factors, that when multiplied together give the original polynomial. There are several techniques for factoring:
  • **Identifying patterns:** like perfect square trinomials or the difference of squares.
  • **Grouping terms:** useful when dealing with larger polynomials.
  • **Trial and error:** finding pairs of factors that work with given quadratics.

For the exercise polynomial \(x^2 + 10x + 25\), recognizing it as a perfect square trinomial allows us to directly write it as \((x+5)^2\). Mastery of these techniques can significantly enhance problem-solving skills in algebra.

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Most popular questions from this chapter

The binomial \(x^{6}-y^{6}\) may be considered as either a difference of squares or a difference of cubes. Factor \(x^{6}-y^{6}\) by first factoring as a difference of cubes. Then factor further by considering one of the factors as a difference of squares.

The following exercises are of mixed variety. Factor each polynomial. $$ 21 m^{4}-32 m^{2}-5 $$

The following exercises are of mixed variety. Factor each polynomial. $$ x^{5}+3 x^{4}-x-3 $$

Factor each polynomial. $$ 64 g^{3}-27 h^{3} $$

Factor each polynomial. $$ 27 y^{3}+1 $$
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