Problem 14 - Precalculus

Question

Express each polynomial in the form \(a_{n}\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{n}\right)\). $$6 x^{2}+x-12$$

Step-by-Step Solution

Answer

Factored form: \((3x-4)(2x+3)\).

Step 1: Write Down the Polynomial

The given polynomial is \(6x^2 + x - 12\). Our goal is to express it in factored form.

Step 2: Identify a and c

Identify the coefficients \(a = 6\) (for \(x^2\)) and \(c = -12\). These will be used to find the pair of numbers that multiply to \(ac\) and add to the middle term.

Step 3: Find Product ac

Calculate the product \(ac = 6 \times (-12) = -72\).

Step 4: Factor ac into Two Numbers

We need to find two numbers that multiply to \(-72\) and add to \(1\) (the coefficient of \(x\)). These numbers are \(9\) and \(-8\), because \(9 imes -8 = -72\) and \(9 + (-8) = 1\).

Step 5: Rewrite the Middle Term

Rewrite the polynomial as \(6x^2 + 9x - 8x - 12\). We have split the middle term \(x\) into two terms using our pair: \(9x\) and \(-8x\).

Step 6: Group and Factor by Grouping

Group the terms: \((6x^2 + 9x) + (-8x - 12)\). Factor out the greatest common factor (GCF) from each group: \(3x(2x + 3) - 4(2x + 3)\).

Step 7: Factor Out the Common Binomial

Factor out \((2x + 3)\): \((3x - 4)(2x + 3)\).

Step 8: Write the Result in Specified Form

The expression \(6x^2 + x - 12\) in the form \(a_n(x - r_1)(x - r_2)\) is \((3x - 4)(2x + 3)\). Here, \(a_n = 1\), \(r_1 = -\frac{3}{2}\), \(r_2 = \frac{4}{3}\).

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation that takes the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). This is called a quadratic because the highest degree of the variable \(x\) is squared. Quadratic equations can describe simple shapes like parabolas when graphed. They appear frequently in various applications ranging from physics to finance. When solving quadratic equations, one of the main goals is to find their roots or solutions — the values of \(x\) that make the equation true. You can solve quadratic equations using different methods:

• Factoring: This method involves expressing the quadratic equation as a product of two binomials.
• Quadratic Formula: This formula, \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), gives exact roots.
• Completing the Square: This involves rewriting the quadratic in a form that makes it easier to solve.
• Graphing: By plotting the equation, you can visually identify the roots.

Factoring by Grouping

Factoring by grouping is a powerful algebraic technique used to factor polynomials, especially when dealing with trinomials or more complex polynomials. This method is very useful when direct factoring seems complicated.The main idea is to rearrange and group terms in the polynomial so that you can factor out a common factor in smaller segments. Here's how it generally works:

• Arrange the Polynomial: Split the middle term if necessary to facilitate grouping.
• Group the Terms: Pair the terms, typically in twos, such that each group has a common factor.
• Factor Within the Groups: Factor out the greatest common factor from each group.
• Factor the Common Binomial: Pull out a common binomial factor, if it exists, which simplifies the expression into a product of two simpler terms.

In the given example, \(6x^2 + 9x - 8x - 12\) is grouped into \((6x^2 + 9x) + (-8x - 12)\). Then, \(3x\) and \(-4\) are factored out of each group leading to a common binomial \((2x + 3)\). This results in the factored form \((3x - 4)(2x + 3)\). This method turns a complex polynomial into a product of simpler factors, which can make calculations and further algebra simpler.

Roots of Polynomial

The roots of a polynomial, also known as zeros, are the values of \(x\) that make the polynomial equal to zero. Discovering these values is essential in solving polynomial equations, as they indicate where the polynomial intersects the x-axis on a graph.To find the roots of a polynomial, you can use various methods. Once factored, as in the solution \((3x - 4)(2x + 3)\), finding the roots becomes straightforward:

• Set each factor equal to zero.
• Solve for \(x\) in each equation.

In our example, setting \((3x - 4) = 0\) gives \(x = \frac{4}{3}\) and setting \((2x + 3) = 0\) gives \(x = -\frac{3}{2}\). These are the roots of the polynomial. Finding these roots means we have effectively identified the points where the quadratic equation crosses the x-axis. This not only provides the solutions to the equation but also offers insights into the behavior of the graph associated with the polynomial. Understanding roots in this factored context simplifies many mathematical problems, revealing critical information about the polynomial function's nature.