Problem 52
Question
Factor any perfect square trinomials, or state that the polynomial is prime. $$x^{2}+24 x+144$$
Step-by-Step Solution
Verified Answer
The factored form of the perfect square trinomial \(x^2 + 24x + 144\) is \((x+12)^2\).
1Step 1: Identify the Coefficients
The given polynomial \(x^2 + 24x + 144\) has coefficients a=1, b=24 and c=144 as per the standard quadratic equation \(ax^2 + bx + c\).
2Step 2: Check for Perfect Square Trinomial
To check if the trinomial is a perfect square, we check if the square root of the first coefficient 'a' and the last coefficient 'c' are equal to half of the middle coefficient 'b'. So we express b = \(2*sqrt(a*c)\). The middle coefficient of the trinomial is 24, which can be expressed as \(2*sqrt(1*144)\), confirming the trinomial is a perfect square.
3Step 3: Factor the Polynomial
We can now write the given trinomial as \(x^2 + 2*12x + 12^2\), which is in the structure \((ax+b)^2 = a^2x^2+2abx+b^2\). With this, we can simplify the trinomial into the perfect square binomial form \((x+12)^2\).
Key Concepts
Perfect Square TrinomialsQuadratic EquationsAlgebraic Expressions
Perfect Square Trinomials
Perfect square trinomials are special types of quadratic expressions that take the form \( (ax+b)^2 \) or \( (ax-b)^2 \). These trinomials can be expanded into \( a^2x^2 \pm 2abx + b^2 \). To determine if a trinomial is a perfect square, you'll want to look for these characteristics:
- The first and last terms are perfect squares themselves.
- The middle term can be expressed as twice the product of the square roots of the first and last terms.
Quadratic Equations
Quadratic equations are algebraic expressions that involve a variable raised to the power of 2. They are usually written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). Solving these equations often involves factoring the quadratic expression, completing the square, or using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Factoring is often the simplest method, particularly when the quadratic expression is a perfect square trinomial. In our example of \( x^2 + 24x + 144 \), recognizing it as a perfect square trinomial allows us to factor it as \( (x+12)^2 \), which simplifies the solving process.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations. They can take many forms, such as polynomials, rational expressions, or radical expressions. Understanding how to manipulate and simplify these expressions is key to solving algebraic problems.
For instance, a polynomial is a type of algebraic expression that includes terms such as \( ax^n \), where \( n \) is a non-negative integer. In our context, when we factor expressions like \( x^2 + 24x + 144 \), we are transforming them into simpler or more useful forms, such as the perfect square binomial \( (x+12)^2 \).
This transformation process involves breaking down complex expressions into components that are easier to handle, offering clearer insights into their structure and potential solutions.
For instance, a polynomial is a type of algebraic expression that includes terms such as \( ax^n \), where \( n \) is a non-negative integer. In our context, when we factor expressions like \( x^2 + 24x + 144 \), we are transforming them into simpler or more useful forms, such as the perfect square binomial \( (x+12)^2 \).
This transformation process involves breaking down complex expressions into components that are easier to handle, offering clearer insights into their structure and potential solutions.
Other exercises in this chapter
Problem 52
Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$y(y+9)=4(2 y+5)$$
View solution Problem 52
Factor each polynomial using the negative of the greatest common factor. $$-18 x^{4}+9 x^{3}+6 x^{2}$$
View solution Problem 52
Use the method of your choice to factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$15 x^{2}-31
View solution Problem 53
Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations usin
View solution