Problem 31 - Precalculus : Building Concepts and Connections

Question

Determine together \(q(x)\) is a factor of \(p(x)\) Here, \(p(x)\) is the first polynomial and \(q(x)\) is the second polynomial. justify your answer. $$x^{3}-7 x+6 ; x+3$$

Step-by-Step Solution

Answer

Yes, \(q(x) = x+3\) is a factor of polynomial \(p(x) = x^{3}-7x+6\).

Step 1: Identify the 'a' value

First, find the 'a' value from the polynomial \(q(x)\). In this case, \(q(x) = x+3\) (which is in the form \(x-a\)). Therefore, the value of 'a' in this case would be '-3'.

Step 2: Substitute 'a' into \(p(x)\)

Next, you need to substitute 'a' into \(p(x)\). So we substitute -3 into \(p(x)\), which gives \(p(-3)\). This equals to \((-3)^{3}-7*(-3)+6\).

Step 3: Solve the equation

Solving the equation \((-3)^{3}-7*(-3)+6\), it simplifies to \(-27+21+6\), which is equal to 0.

Step 4: Verify factor

According to the factor theorem, if a polynomial \(p(a)\) equals 0, then \(x-a\) is a factor of that polynomial. Since \(p(-3) = 0\), \(x+3\) is a factor of \(p(x)\).

Key Concepts

Polynomials

Polynomials are mathematical expressions consisting of variables and coefficients that are combined using only addition, subtraction, multiplication, and non-negative integer exponents. For example, the polynomial in the exercise, \(p(x) = x^3 - 7x + 6\), is composed of three terms: \(x^3\), \( -7x\), and \(6\). The highest exponent in a polynomial is known as the degree of the polynomial; in our case, the degree is 3 because the highest exponent is 3.

Understanding polynomials is crucial because they're the backbone of algebra and appear in many different areas of mathematics and sciences. They are used to model a wide variety of real-world situations, from simple physics problems to complex financial calculations.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler 'factor' polynomials that, when multiplied together, will result in the original polynomial. For instance, when we take \(x^3 - 7x + 6\) and factor it, we may get \(x - 1\), \(x + 3\), and another factor not identified in the exercise.

Factoring is akin to finding the 'building blocks' of a polynomial. These blocks, or factors, are often much simpler to work with, especially when solving equations. A key point in factoring is that not all polynomials can be factored into linear factors, which involve only first-degree terms like \(x-a\). Nonetheless, factoring when it is possible can greatly simplify solving polynomial equations.

Substitution Method

The substitution method is used in algebra to replaces a variable with a given number or another expression. It's a crucial approach for evaluating polynomials at a particular point, determining the value of a polynomial function, or simplifying complex algebraic expressions.

In our exercise, we substituted the value \( -3\) for \(x\) in the polynomial \(p(x)\) to evaluate \(p(-3)\). By using this method, we can test if \(x+3\) is a factor of \(p(x)\) by checking if the polynomial evaluates to zero when \(x\) is replaced by \( -3\). The substitution method is not only foundational for understanding polynomials but also for many other areas of mathematics.

Solving Polynomial Equations

Solving polynomial equations involves finding all the values (roots) that make the equation true. When a polynomial is set equal to zero, these roots are the solutions to the equation. For example, finding the roots of the polynomial equation \(p(x) = 0\) would entail unveiling the values of \(x\) that satisfy the equation.

The Factor Theorem is particularly handy in solving polynomial equations, saying that if \(p(a) = 0\), then \(x - a\) is a factor of the polynomial \(p(x)\). This theorem guided us in the exercise to conclude that because \(p(-3)\) equals zero, the binomial \(x+3\) is indeed a factor of \(p(x)\). Solving polynomial equations is vital in various disciplines, including engineering, economics, and physics, as it helps describe and solve real-life situations.