Problem 7
Question
Factor completely, by hand or by calculator. Check your results. Trinomials with a Leading Coefficient of 1. $$x^{2}+7 x+12$$
Step-by-Step Solution
Verified Answer
The factored form is \((x + 3)(x + 4)\).
1Step 1: Identify the Trinomial
The given trinomial is a quadratic in the form of \(ax^{2} + bx + c\) where \(a = 1\), \(b = 7\), and \(c = 12\). Since \(a = 1\), we are dealing with a simple case of factoring.
2Step 2: Find Two Numbers
We need to find two numbers that multiply to \(ac\) (which is simply \(c\) since \(a=1\)) and add up to \(b\). In this case, we look for two numbers that multiply to \(12\) and add up to \(7\).
3Step 3: Trial and Error
By trial and error, we find that the numbers 3 and 4 multiply to 12 \((3 \times 4 = 12)\) and add up to 7 \((3 + 4 = 7)\).
4Step 4: Write the Factors
The trinomial can be factored as \((x + 3)(x + 4)\), which are the factors of the quadratic equation \(x^{2}+7x+12\).
5Step 5: Verify the Factors
To check, expand the factors: \((x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12\), which matches the original trinomial.
Key Concepts
Trinomials with a Leading Coefficient of 1Factoring TrinomialsQuadratic Equations
Trinomials with a Leading Coefficient of 1
Understanding how to factor trinomials is a fundamental skill in algebra, particularly when solving quadratic equations. A trinomial with a leading coefficient of 1 is a form of quadratic equation expressed as \(x^2+bx+c\). The leading coefficient is the number before the \(x^2\) term, and in this case, it's 1.
The simplicity of having a leading coefficient of 1 makes the factoring process more straightforward. The goal is to find two binomials that, when multiplied together, produce the original trinomial. To factor such trinomials, we look for two numbers whose product is equal to the constant term c, and whose sum is equal to the coefficient b of the x term. These two numbers can then be used to write down the factors of the trinomial as \((x + m)(x + n)\), where m and n are the numbers you have discovered.
Factoring becomes an invaluable tool for solving a wide range of problems in algebra, from simplifying expressions to finding zeros of functions. It's an essential step that enables the student to progress to more complex algebraic manipulations.
The simplicity of having a leading coefficient of 1 makes the factoring process more straightforward. The goal is to find two binomials that, when multiplied together, produce the original trinomial. To factor such trinomials, we look for two numbers whose product is equal to the constant term c, and whose sum is equal to the coefficient b of the x term. These two numbers can then be used to write down the factors of the trinomial as \((x + m)(x + n)\), where m and n are the numbers you have discovered.
Factoring becomes an invaluable tool for solving a wide range of problems in algebra, from simplifying expressions to finding zeros of functions. It's an essential step that enables the student to progress to more complex algebraic manipulations.
Factoring Trinomials
Factoring trinomials involves rewriting a trinomial as the product of two or more simpler polynomials. The trinomials we often start learning to factor are in the form of \(x^2 + bx + c\), where \(a=1\). Factoring is essentially the reverse of multiplying polynomials.
The process usually involves a bit of trial and error to find two numbers that meet the criteria for multiplication and addition. An easy way to approach this is by creating a list of pairs of factors of the constant term c and testing which pair adds up to the middle coefficient, b. Once the correct pair is found, like in the example exercise where 3 and 4 are the magical numbers for 12 and 7 respectively, the trinomial can be broken down into \((x + 3)(x + 4)\).
Understanding this process not only aids in algebra but also builds foundational skills for higher-level mathematics including calculus, where factoring can be crucial for simplification and solving equations.
The process usually involves a bit of trial and error to find two numbers that meet the criteria for multiplication and addition. An easy way to approach this is by creating a list of pairs of factors of the constant term c and testing which pair adds up to the middle coefficient, b. Once the correct pair is found, like in the example exercise where 3 and 4 are the magical numbers for 12 and 7 respectively, the trinomial can be broken down into \((x + 3)(x + 4)\).
Understanding this process not only aids in algebra but also builds foundational skills for higher-level mathematics including calculus, where factoring can be crucial for simplification and solving equations.
Quadratic Equations
Quadratic equations are second-order polynomials with three terms, typically written in the form of \(ax^2 + bx + c = 0\). The solutions to these equations are found by factoring, completing the square, using the quadratic formula, or graphing.
In our textbook problem, the quadratic was present in its factorable form with a leading coefficient of 1. Factoring is often the quickest method to solve when the quadratics are neat, with small integer solutions. By setting each binomial factor to zero, you can solve for the 'x' values, often called the zeros or roots of the equation. For instance, by factoring \(x^2 + 7x + 12\), we obtain the roots -3 and -4. These roots are the answers to the equation \(x^2 + 7x + 12 = 0\).
A strong grasp of factoring quadratics is pivotal in understanding the behavior of parabolas, which is the shape described by graphing a quadratic equation. Beyond just algebra, they have practical applications in physics, engineering, economics, and various other sciences.
In our textbook problem, the quadratic was present in its factorable form with a leading coefficient of 1. Factoring is often the quickest method to solve when the quadratics are neat, with small integer solutions. By setting each binomial factor to zero, you can solve for the 'x' values, often called the zeros or roots of the equation. For instance, by factoring \(x^2 + 7x + 12\), we obtain the roots -3 and -4. These roots are the answers to the equation \(x^2 + 7x + 12 = 0\).
A strong grasp of factoring quadratics is pivotal in understanding the behavior of parabolas, which is the shape described by graphing a quadratic equation. Beyond just algebra, they have practical applications in physics, engineering, economics, and various other sciences.
Other exercises in this chapter
Problem 7
Simplify. Leave your answers as improper fractions. $$\frac{x+\frac{y}{4}}{x-\frac{y}{3}}$$
View solution Problem 7
Factor completely.$$x^{2}-9 y^{2}$$
View solution Problem 7
Simplify each fraction by manipulating the algebraic signs. $$\frac{a-b}{b-a}$$
View solution Problem 8
Multiply and reduce. Do some by calculator. $$\frac{3}{5} \times \frac{2}{7} \times \frac{5}{9}$$
View solution