Problem 59
Question
Factor completely. $$4 x^{2}+26 x+30$$
Step-by-Step Solution
Verified Answer
The quadratic \(4 x^{2}+26 x+30\) completely factors to \( (2x+5)(2x+6) \)
1Step 1: Set up the quadratic
We have a quadratic of the form \( ax^2 + bx + c \) where \( a = 4 \), \( b = 26 \), and \( c = 30 \). Our aim is to find two numbers whose product is \( ac = 4*30 =120 \) and sum is \( b = 26 \).
2Step 2: Identify the numbers
The numbers that satisfy these conditions are 10 and 12 since \( 10*12 = 120 \) and \( 10+12 = 22 \).
3Step 3: Rewrite the equation
We use these two numbers to rewrite the middle term of the equation. The equation becomes: \( 4x^2 + 10x + 16x + 30 \).
4Step 4: Factor by grouping
Arrange the terms so that we can factor by grouping. The equation can be rewritten as \( (4x^2 +10x) + (16x + 30) \). Factoring gives us \( 2x(2x+5) + 2(8x+15) \).
5Step 5: Checking and final result
We notice that \( (2x+5) \) is a common factor. Factoring out \( (2x+5) \) gives us the final factored form of the equation: \( (2x+5)(2x+6) \) as the final answer.
Key Concepts
Factoring by GroupingQuadratic EquationsPolynomial Factorization
Factoring by Grouping
Factoring by grouping is a strategy used to simplify polynomials by organizing terms into smaller groups that can be factored separately. This method proves particularly valuable when dealing with four-term polynomials or when a common factor is not immediately apparent.
In this process, you start by dividing your polynomial into two groups. Each group should have two terms. In our exercise, the quadratic expression was rewritten as four terms: \( 4x^2 + 10x + 16x + 30 \).
Once grouped, factor out the greatest common factor (GCF) from each pair.
For example:
In this process, you start by dividing your polynomial into two groups. Each group should have two terms. In our exercise, the quadratic expression was rewritten as four terms: \( 4x^2 + 10x + 16x + 30 \).
Once grouped, factor out the greatest common factor (GCF) from each pair.
For example:
- First group: \( 4x^2 + 10x \) can be factored as \( 2x(2x + 5) \).
- Second group: \( 16x + 30 \) simplifies to \( 2(8x + 15) \).
Quadratic Equations
Quadratic equations are expressions of the form \( ax^2 + bx + c = 0 \). These equations involve a term in \( x^2 \) and typically represent parabolas when graphed. The structure is particularly known for its application in many areas, from physics to finance.
In the given problem, the quadratic expression is \( 4x^2 + 26x + 30 \). Here, \( a = 4 \), \( b = 26 \), and \( c = 30 \). Understanding these components is key to managing quadratic equations effectively.
Quadratic equations can be solved in several ways, such as:
In the given problem, the quadratic expression is \( 4x^2 + 26x + 30 \). Here, \( a = 4 \), \( b = 26 \), and \( c = 30 \). Understanding these components is key to managing quadratic equations effectively.
Quadratic equations can be solved in several ways, such as:
- Factoring
- Completing the square
- Quadratic formula
Polynomial Factorization
Polynomial factorization is the process of expressing a polynomial as the product of simpler polynomials. It's a useful skill in algebra for simplifying expressions and solving equations, especially when dealing with higher-degree polynomials.
To factor a polynomial like \( 4x^2 + 26x + 30 \), first identify the general form and calculate the product \( ac \) of the first and last coefficients. In this case, it's \( 120 \). The goal is to find two numbers that multiply to \( 120 \) and add to the middle coefficient \( b = 26 \). The numbers \( 10 \) and \( 12 \) meet these requirements.
This step helps in rewriting and breaking down the polynomial into groups, enabling the factoring by grouping method. Successfully factoring a polynomial requires careful attention to detail and practice, but it's an indispensable tool for handling complex algebraic expressions.
To factor a polynomial like \( 4x^2 + 26x + 30 \), first identify the general form and calculate the product \( ac \) of the first and last coefficients. In this case, it's \( 120 \). The goal is to find two numbers that multiply to \( 120 \) and add to the middle coefficient \( b = 26 \). The numbers \( 10 \) and \( 12 \) meet these requirements.
This step helps in rewriting and breaking down the polynomial into groups, enabling the factoring by grouping method. Successfully factoring a polynomial requires careful attention to detail and practice, but it's an indispensable tool for handling complex algebraic expressions.
Other exercises in this chapter
Problem 59
Factor each polynomial using the greatest common binomial factor. $$x(x+2)-4(x+2)$$
View solution Problem 59
Factor any perfect square trinomials, or state that the polynomial is prime. $$x^{2}+14 x y+49 y^{2}$$
View solution Problem 60
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 Problem 60
Factor completely. $$x^{4}-22 x^{3}+120 x^{2}$$
View solution