Problem 72
Question
Factor. $$ r^{2}+24 r+144 $$
Step-by-Step Solution
Verified Answer
The expression \( r^2 + 24r + 144 \) factors to \((r + 12)^2\).
1Step 1: Identify the quadratic form
We begin with the quadratic expression, which is in the form of \( r^2 + br + c \). In this exercise, it is given as \( r^2 + 24r + 144 \).
2Step 2: Set up for factoring
We need to identify two numbers that multiply to \( c = 144 \) and add up to \( b = 24 \).
3Step 3: Find the correct pair of numbers
By inspecting, we find the numbers 12 and 12 satisfy both conditions: 12 multiplied by 12 equals 144 and added together equals 24.
4Step 4: Write the expression as a perfect square trinomial
Since both factors are the same, we can express the quadratic as a perfect square: \((r + 12)(r + 12)\) or \((r + 12)^2\).
5Step 5: Verify the factorization
Expand \((r + 12)^2\) to check if it returns the original quadratic: \((r + 12)(r + 12) = r^2 + 12r + 12r + 144 = r^2 + 24r + 144\). The factorization is correct.
Key Concepts
Perfect Square TrinomialFactoring TechniquesQuadratic Expressions
Perfect Square Trinomial
A perfect square trinomial is a specific form of quadratic expression that allows for straightforward factorization. It has certain characteristics that make it unique among other quadratics. In essence, when you have a trinomial of the form \[ a^2 + 2ab + b^2 \]it can be expressed neatly as the square of a binomial:\[ (a + b)^2. \]Consider \[ r^2 + 24r + 144 \] from our example. Here, it fits the perfect square trinomial pattern when we recognize it as \[ (r + 12)(r + 12) = (r + 12)^2. \]
- The key here is to find the middle term that equals twice the product of the numbers used in the binomial.
- This means the linear coefficient (\[2ab\]) is 2 times 12 and 12 in this instance.
Factoring Techniques
Factoring techniques are essential tools in algebra, especially when dealing with quadratic equations. They simplify complex expressions, making them easier to solve or manipulate. The most common techniques include:
- Perfect Square Trinomial: Applies when the expression forms a perfect square, as previously discussed. Identify pairs that add and multiply to the correct coefficients.
- Factoring by Grouping: Useful for more complicated quadratics not in perfect square format. It requires splitting the middle term to form groups that can be individually factored.
Quadratic Expressions
Quadratic expressions are algebraic expressions where the highest degree is two. They have a general form:\[ ax^2 + bx + c \]These expressions often describe a parabolic curve when graphed. Understanding quadratics is critical as they appear frequently in various real-world problems. Key features include:
- Leading Coefficient: The term \( a \)determines the direction and width of the parabola.
- Discriminant: Found using \( b^2 - 4ac \), it helps predict the nature of the roots (real and distinct, real and equal, or complex). However, our example \( r^2 + 24r + 144 \) did not require this since we were factoring.
Other exercises in this chapter
Problem 71
Factor each polynomial by factoring out the opposite of the GCF. $$ -4 a^{3} b^{2}+14 a^{2} b^{2}-10 a b^{2} $$
View solution Problem 72
The following is a list of random factoring problems. Factor each expression. If an expression is not factorable, write "prime." See Examples 1-5. $$ 16 d^{2}-5
View solution Problem 72
Choose the correct method from Section 6.1 or Section 6.2 to factor each of the following. $$t^{2}-5 t-50$$
View solution Problem 72
Solve each equation. $$ h(3 h-4)(h+1)=0 $$
View solution