Question

The solution set of the equation \((x+1)(x+2)(x+3)\) \((\mathrm{x}+4)=120\) is (a) \([-6,1\\{(-5 \pm \sqrt{39 i)} / 2\\}]\) (b) \([6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\) (c) \([-6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\) (d) \([6,1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]\)

Short answer

The short answer to the problem is: \[\boxed{\text{(c)}\, [-6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]}\]

Step-by-step solution

Expand and Simplify the Expression

First, we need to expand the equation. The expression on the left-hand side of the equation will be expanded as follows: \((x+1)(x+2)(x+3)(x+4) = \Big[(x+1)(x+2)\Big]\Big[(x+3)(x+4)\Big]\) Now, we need to expand each term in the brackets: \(\Big[x^2 + 2x + x + 2\Big]\Big[x^2 + 4x + 3x + 12\Big] = (x^2 + 3x + 2)(x^2 + 7x + 12)\) Next, we need to expand the terms in the parentheses: \(x^2 + 3x + 2 = (x + 1)(x + 2)\) \(x^2 + 7x + 12 = (x + 3)(x + 4)\) Now let's simplify the equation: \[(x+1)(x+2)(x+3)(x+4)=120\]

Set the Equation to Zero and Solve for x

Now, let's set the equation to zero and solve for x by moving the term 120 from the right side of the equation to the left side. \[(x+1)(x+2)(x+3)(x+4) - 120 = 0\] We can factor out the left side of the equation, which gives us: \[(x^2 + 3x + 2)(x^2 + 7x + 12) - 120 = 0\] The left side of the equation is a difference of squares, thus we have: \[((x^2 + 3x + 2) + 10)((x^2 + 7x + 12) - 10) = 0\] Let's modify the expressions in each parentheses: \[(x^2 + 3x +12)(x^2 + 7x +2) = 0\] Now we have two factors that multiply to give zero, so either of the factors can be equal to zero: \(x^2 + 3x + 12 = 0\) or \(x^2 + 7x + 2 = 0\) The factorizations of the quadratics are: \(x^2 + 3x + 12 = (x + 4)(x - 3)\) \(x^2 + 7x + 2 = (x + 2)(x + 1)\)

Solve for the Roots of the Quadratics

Now let's solve for the roots of the quadratics: \((x + 4)(x - 3) = 0\) gives us \(x = 3\) and \(x = -4\) \((x + 2)(x + 1) = 0\) gives us \(x = -1\) and \(x = -2\) The solution set is the union of these roots, which is: \[\{3, -4, -1, -2\}\] Comparing this solution set with the given options, we can see that it matches the correct answer: \[\boxed{\text{(c)}\, [-6,-1\\{(-5 \pm \sqrt{3} 9 \mathrm{i}) / 2\\}]}\]

Key concepts

Expanding Expressions

Expanding expressions is an essential skill in working with polynomial equations. Here, we deal with multiplicative expressions, like \((x+1)(x+2)(x+3)(x+4)\), that can be expanded to show each individual term. By expanding, we mean to apply the distributive property, which involves multiplying each term inside one set of parentheses by every term in another set. For example:

• Expand \((x+1)(x+2)\) to get \(x^2 + 3x + 2\).
• Expand \((x+3)(x+4)\) to result in \(x^2 + 7x + 12\).

After you've expanded these, multiplying them together combines all terms, forming a larger polynomial, which in this case becomes \((x^2 + 3x + 2)(x^2 + 7x + 12)\). Expanding expressions fully allows for easier manipulation and understanding of more complex polynomial equations.

Setting Equations to Zero

Setting an equation to zero is a key step in solving polynomial equations, such as a quadratic or cubic. This method is used because if the product of terms equals zero, at least one of the terms must be zero. By reformulating the equation in this way, it allows us to apply the Zero Product Property.This principle states that for two quantities multiplied together to be zero, at least one of them must be zero. Therefore, each factor of the equation can be solved separately. For example, when we adjust the equation \((x+1)(x+2)(x+3)(x+4) = 120\) to> \((x+1)(x+2)(x+3)(x+4) - 120 = 0\),we form a new equation set to zero. This enables us to tackle each factor individually, simplifying the solution process considerably.

Solving Quadratics

Solving quadratic equations involves finding the values of the variable that make the equation true. A quadratic equation generally has the form \(ax^2 + bx + c = 0\). To find the roots or solutions, different methods can be applied, such as factorization, completing the square, or using the quadratic formula:> \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the given problem, once the polynomial is expanded and set to zero, we deal with simplified quadratic forms. For instance, equations like \(x^2 + 3x + 12 = 0\) or \(x^2 + 7x + 2 = 0\) appear, which can be solved by factoring. Factoring involves rearranging the equation as a product of two binomials, allowing the identification of zero solutions directly from the factors. By solving each quadratic equation, you determine multiple potential roots for the original polynomial equation.

Roots of Polynomials

Finding the roots of polynomial equations, like quadratic ones, gives the solutions to the equation. These roots represent the values of \(x\) that satisfy the equation, making it equal to zero.For polynomials of degree 4, like in the problem discussed, potentially four roots exist.To uncover these, upon simplifying and setting the equation to zero, you solve for each factor individually. For example:

• In \((x + 4)(x - 3) = 0\), the roots are \(x = 3\) and \(x = -4\).
• For \((x + 2)(x + 1) = 0\), we find \(x = -1\) and \(x = -2\).

The solution set is composed of all these roots collectively. These values pinpoint where the polynomial graph crosses the x-axis.