Problem 21
Question
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 using multiplication or a graphing utility. $$5 x^{2}-5 x-30$$
Step-by-Step Solution
Verified Answer
So the complete factorization of the polynomial \(5 x^{2}-5 x-30\) is \(5(x - 3)(x + 2)\).
1Step 1: Express the Polynomial into a Suitable Form
The first step is to express the given polynomial into an easily factorable form. Since all terms in the polynomial have \(5\) as a factor, factor out \(5\). This gives, \(5(x^{2} - x - 6)\).
2Step 2: Further Factoring using Grouping
To factorise the expression \(x^{2} - x - 6\) further, we can use factoring by grouping. We first look for two numbers whose product is \(-6\) (the last term) and whose sum is \(-1\) (the coefficient of the second term). These numbers are \(-3\) and \(2\). So, the polynomial can be expressed as \(5(x^{2} - 3x + 2x - 6)\). This can be further factored by grouping to get \(5[(x - 3)(x + 2)]\).
3Step 3: Check Factoring by Multiplication
We can check the factoring by multiplying out the factors we obtain. We get \(5*x*x-15x+10*x-30 = 5x^{2}-5x-30\), which is the original polynomial.
Key Concepts
Factoring TechniquesGrouping MethodChecking FactorizationQuadratic Expressions
Factoring Techniques
Factoring techniques are crucial for simplifying polynomials and making them easier to work with. There are various methods to factor a polynomial, depending on its form and terms. In this exercise, we're focusing on a common technique: factoring out a common factor. This is often the first step because it's a simple way to reduce the complexity of the polynomial.
Essentially, you identify any common numerical factor present in all the terms. For example, in the polynomial given, each term includes the factor 5. By factoring out 5, the polynomial becomes much simpler:
Essentially, you identify any common numerical factor present in all the terms. For example, in the polynomial given, each term includes the factor 5. By factoring out 5, the polynomial becomes much simpler:
- Given: \(5x^2 - 5x - 30\)
- Factor out 5: \(5(x^2 - x - 6)\)
Grouping Method
The grouping method is a technique used to factor polynomials, especially useful when dealing with quadratics that can be tricky to factor at first glance. After factoring out a common factor, if the remaining expression is not obvious to factor, we can apply the grouping method.
Here's how it works for the quadratic part of our example: \(x^2 - x - 6\). The idea is to split the middle term in such a way that the expression can then be reshaped into groups that are easier to manage:
Here's how it works for the quadratic part of our example: \(x^2 - x - 6\). The idea is to split the middle term in such a way that the expression can then be reshaped into groups that are easier to manage:
- Identify pairs of terms whose product equals the last term (constant), and whose sum equals the middle term's coefficient:
- Product needed: \(-6\)
- Sum needed: \(-1\)
- Numbers found: \(-3\) and \(2\)
- Rewrite: \(x^2 - x - 6\) as \(x^2 - 3x + 2x - 6\)
- Group: \((x^2 - 3x) + (2x - 6)\)
- Factor: \(x(x - 3) + 2(x - 3)\)
- Combine: \((x + 2)(x - 3)\)
Checking Factorization
Once you've factored a polynomial, it's essential to verify your work. This ensures that no mistakes were made in the factoring process. There are different methods to check your factorization, with multiplication being one of the most straightforward approaches.
To check the factorization of our example:
To check the factorization of our example:
- Factors obtained: \(5(x - 3)(x + 2)\)
- Expand:
- First, expand \((x - 3)(x + 2) = x^2 + 2x - 3x - 6 = x^2 - x - 6\)
- Then multiply by the factor 5: \(5(x^2 - x - 6)\) = \(5x^2 - 5x - 30\)
Quadratic Expressions
Quadratic expressions are a common type of polynomial expression and have a standard form: \(ax^2 + bx + c\). They're defined by the terms' combination, with \(x^2\) being the highest power of x known as the quadratic term. Understanding how to factor these is essential in algebra, as it helps simplify complex equations and find roots or solutions.
In our given exercise, the expression inside the parentheses after factoring out the common factor 5 is a quadratic expression: \(x^2 - x - 6\). Working through this involves solving it via:
In our given exercise, the expression inside the parentheses after factoring out the common factor 5 is a quadratic expression: \(x^2 - x - 6\). Working through this involves solving it via:
- Identifying suitable numbers for splitting the middle term, which aids in factorization.
- Applying methods such as the grouping method, contributing to making it factorable.
Other exercises in this chapter
Problem 21
Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$x^{2}=4 x$$
View solution Problem 21
Factor each difference of two squares. $$x^{4}-y^{10}$$
View solution Problem 21
Factor each polynomial using the greatest common factor. If there is no common factor other than 1 and the polynomial cannot be factored, so state. $$x^{2}+5 x$
View solution Problem 21
Use the method of your choice to factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$8 x^{2}+33 x
View solution