Problem 110 - Intermediate Algebra

Question

Let \(f(x)=\frac{x^{3}+4 x^{2}}{25 x+100} .\) For what values of \(x\) is \(f(x)=1 ?\)

Step-by-Step Solution

Answer

The values of \(x\) are 5, -4, and -5.

Step 1: Set up the equation

We need to solve the equation \(f(x) = 1\). The function \(f(x)\) is given by \(f(x) = \frac{x^3 + 4x^2}{25x + 100}\). Set this equal to 1: \[\frac{x^3 + 4x^2}{25x + 100} = 1.\]

Step 2: Eliminate the fraction

To eliminate the fraction, multiply both sides of the equation by the denominator \(25x + 100\):\[(x^3 + 4x^2) = 1 \cdot (25x + 100)\]. This simplifies to \[x^3 + 4x^2 = 25x + 100.\]

Step 3: Rearrange the equation

Rearrange the equation by bringing all terms to one side to set the equation to zero:\[x^3 + 4x^2 - 25x - 100 = 0.\]

Step 4: Solve the cubic equation

We need to solve the cubic equation \(x^3 + 4x^2 - 25x - 100 = 0\). Let's try to factor it by looking for rational roots using the Rational Root Theorem. Possible rational roots are divisors of the constant term (-100), which are \(\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20, \pm 25, \pm 50, \pm 100\).

Step 5: Test possible roots

Test these possible roots in the equation. For instance, try \(x = 5\):\[5^3 + 4(5^2) - 25(5) - 100 = 125 + 100 - 125 - 100 = 0.\]Thus, \(x = 5\) is a root.

Step 6: Factor out \(x - 5\)

Since \(x = 5\) is a root, \(x - 5\) is a factor of \(x^3 + 4x^2 - 25x - 100\). Use polynomial division to divide \(x^3 + 4x^2 - 25x - 100\) by \(x - 5\).

Step 7: Complete the polynomial division

Perform the division to find the quotient:Starting with leading terms:\(x^3 \div x = x^2\). Multiply \(x^2\) by \(x - 5\) to get \(x^3 - 5x^2\). Subtract to get \(9x^2 - 25x - 100\).Continue with \(9x^2 \div x = 9x\). Multiply \(9x\) by \(x - 5\) to get \(9x^2 - 45x\). Subtract to get \(20x - 100\).Finally, \(20x \div x = 20\). Multiply \(20\) by \(x - 5\) to get \(20x - 100\). Subtract to get 0, confirming the division.The quotient is \(x^2 + 9x + 20\).

Step 8: Factor the quotient

Factor the quadratic \(x^2 + 9x + 20\):The factors of 20 that add to 9 are 4 and 5; therefore, \(x^2 + 9x + 20 = (x + 4)(x + 5)\).

Step 9: Solve for the other roots

The polynomial \(x^3 + 4x^2 - 25x - 100\) can now be factored as \((x - 5)(x + 4)(x + 5)\). Set each factor to zero: \[(x - 5) = 0 \rightarrow x = 5,\]\[(x + 4) = 0 \rightarrow x = -4,\]\[(x + 5) = 0 \rightarrow x = -5.\]

Step 10: Write the solution

The values of \(x\) that satisfy \(f(x) = 1\) are \(x = 5, -4, \text{ and } -5\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool in solving polynomial equations. This theorem helps us to find potential rational solutions for a polynomial equation. For a polynomial expression of the form:

• \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0\)

it states that any possible rational root, in its reduced form \(\frac{p}{q}\), is such that \(p\) is a factor of the constant term \(a_0\), and \(q\) is a factor of the leading coefficient \(a_n\).In our given cubic equation \(x^3 + 4x^2 - 25x - 100 = 0\), the leading coefficient is 1 and the constant term is -100. Thus, the possible rational roots are the factors of -100. We list these possible roots as:

• \(\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20, \pm 25, \pm 50, \pm 100\)

Checking these candidates one by one in the polynomial, as seen in the solution step, can help us identify actual roots that satisfy the equation.

Polynomial Division

Polynomial division is a method used to simplify division between polynomials, similar to long division with numbers. You use it when you want to divide one polynomial by another, and it's especially helpful when verifying factors or simplifying equations.For the cubic equation \(x^3 + 4x^2 - 25x - 100 = 0\), after identifying one of the roots \(x = 5\), we split the polynomial by dividing it with \(x - 5\). The steps involve:

• Dividing the leading term of the dividend by the leading term of the divisor.
• Using the result to multiply the divisor and subtracting from the original polynomial.
• Repeating the process with the new lower degree polynomial formed after subtraction.

This process is repeated until the remainder is zero, indicating a successful division. The quotient found is another polynomial (here, \(x^2 + 9x + 20\)) that can be further factored to find more roots.

Factoring Polynomials

Factoring polynomials is an essential algebraic skill that helps simplify expressions and solve equations. It involves writing a polynomial as a product of its simpler polynomial factors. This process can be particularly straightforward or require methods such as grouping, using special formulas, or trial and error with factoring strategies.In our polynomial \(x^2 + 9x + 20\), we want to express it as a product of two binomials. We look for two numbers that multiply to the constant term (20) and add up to the coefficient of the linear term (9). These numbers are 4 and 5. Therefore, the factorization is:

• \((x + 4)(x + 5) = x^2 + 9x + 20\).

Once the polynomial is fully factored, we can set each factor equal to zero to solve for the roots of the equation. For the cubic polynomial \(x^3 + 4x^2 - 25x - 100 = 0\), after dividing by \(x - 5\), we solved:

• \((x - 5)(x + 4)(x + 5) = 0\)
• Leading to the solutions: \(x = 5, -4, \text{ and } -5\).

Factoring polynomials breaks down complex equations into more manageable parts, making it easier to find solutions.