Problem 55 - Precalculus Mathematics for Calculus

Question

Use the Factor Theorem to show that \(x-c\) is a factor of \(P(x)\) for the given value(s) of \(c\). $$P(x)=2 x^{3}+7 x^{2}+6 x-5, \quad c=\frac{1}{2}$$

Step-by-Step Solution

Answer

Yes, \(x - \frac{1}{2}\) is a factor of \(P(x)\).

Step 1: Setup Polynomial Division

We need to verify whether \(x - \frac{1}{2}\) is a factor of \(P(x) = 2x^3 + 7x^2 + 6x - 5\). According to the Factor Theorem, for \(x - c\) to be a factor, \(P(c)\) should be equal to zero. Here, \(c = \frac{1}{2}\).

Step 2: Substitute the Value

Substitute \(x = \frac{1}{2}\) into the polynomial to check \(P\left(\frac{1}{2}\right)\). Calculate:\[ P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + 7\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) - 5 \]

Step 3: Simplify the Substitution

Simplify the expression, \[ P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) + 7\left(\frac{1}{4}\right) + 6\left(\frac{1}{2}\right) - 5 \]Calculate each term:1. \(2 \times \frac{1}{8} = \frac{1}{4}\)2. \(7 \times \frac{1}{4} = \frac{7}{4}\)3. \(6 \times \frac{1}{2} = 3\)4. \(-5\)

Step 4: Combine the Terms

Combine the terms calculated:\[ \frac{1}{4} + \frac{7}{4} + 3 - 5 \] Simplify using common denominators:\[ = \frac{1 + 7}{4} + 3 - 5 = \frac{8}{4} + 3 - 5 = 2 + 3 - 5 \] \[ = 5 - 5 = 0 \]

Step 5: Conclusion

Since \(P\left(\frac{1}{2}\right) = 0\), according to the Factor Theorem, \(x - \frac{1}{2}\) is indeed a factor of \(P(x)\).

Key Concepts

Polynomial Division

Polynomial division helps break down complex expressions into simpler components. It is similar to long division with numbers, but we apply it to divide polynomials. In our case, checking if \(x - \frac{1}{2}\) is a factor of \(P(x)\) involves polynomial division. We aim to see if this division results in no remainder.
So, what are the steps involved?

• First, we align our polynomials similarly to aligning numbers in long division.
• Next, we divide the leading term of the dividend by the leading term of the divisor, here starting with \(2x^3\) by \(x\).
• We multiply this term by the divisor and subtract from the dividend.
• Repeat the process with the new polynomial expression.

Each stage reduces the expression until you either finish or are left with a remainder.
If the remainder is zero, the divisor is a factor, fulfilling the main check when using polynomial division.

Polynomial Roots

Polynomial roots are values of \(x\) that make the entire polynomial equal to zero. The Factor Theorem relates directly to finding these roots. Specifically, if \(P(c) = 0\), then \(x-c\) is a root of polynomial \(P(x)\). Finding these roots helps in simplifying the polynomial and solving polynomial equations.
Considering the exercise, \(x = \frac{1}{2}\) could be a root if substituting into the polynomial results in zero. This is a simple trial method to determine roots:

• Identify the value \(c\).
• Substitute it into the polynomial \(P(c)\).
• If it equals zero, confirm \(c\) as a root and \(x - c\) as a factor.

Understanding roots is a crucial step in graphing polynomial functions, as they indicate x-intercepts or points where the graph crosses the x-axis.

Substitution Method

The substitution method is a straightforward technique used to evaluate expressions or check solutions. It involves replacing variables with specific numerical values to simplify and solve equations.Here's how it applies:

• You place the desired value into the polynomial equation.
• You perform arithmetic operations where necessary.
• Analyze if the expression results in zero.

In the given problem, substituting \(x = \frac{1}{2}\) helps determine if \(x - \frac{1}{2}\) is a factor by calculating \(P(\frac{1}{2})\) and seeing if the result equals zero.
Composing each part in our substitution: calculate \(2(\frac{1}{2})^3\), then \(7(\frac{1}{2})^2\), and so on, demonstrated it reduces to zero. The presence of zero confirms \(x - \frac{1}{2}\) as a legitimate factor, showcasing how substitution verifies if a factor fits the polynomial perfectly.