Problem 33 - College Algebra with Corequisite Support

Question

For the following exercises, use the Rational Zero Theorem to find the real solution(s) to each equation. $$ x^{4}-2 x^{3}-7 x^{2}+8 x+12=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = 3, 2, -2, -1 \).

Step 1: Understanding the Rational Zero Theorem

The Rational Zero Theorem states that any rational solution (zero) of a polynomial equation, where the coefficients are integers, is of the form \( \frac{p}{q} \) where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.

Step 2: List Possible Rational Zeros

For the polynomial \( x^4 - 2x^3 - 7x^2 + 8x + 12 = 0 \), the constant term is 12 and the leading coefficient is 1. The possible values for \( p \) (factors of 12) are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \) and for \( q \) (factors of 1) are \( \pm 1 \). Hence, the possible rational zeros are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).

Step 3: Testing Possible Rational Zeros

Start testing the possible rational zeros using synthetic division or direct substitution into the polynomial. Begin with \( x = 1 \):\[ 1^4 - 2(1)^3 - 7(1)^2 + 8(1) + 12 = 1 - 2 - 7 + 8 + 12 = 12 \]1 is not a root because the result is 12, not 0. Experiment with other candidates like \( x = 2, -1, 3, etc. \).

Step 4: Find a Zero Using Synthetic Division

Try \( x = 3 \) using synthetic division:For \( 3 \), the division is:3 | 1 -2 -7 8 12 | 3 3 -12 -12 ------------------- 1 1 -4 -4 0Since the remainder is 0, \( x = 3 \) is a zero. The resulting polynomial is \( x^3 + x^2 - 4x - 4 \).

Step 5: Factor the Reduced Polynomial

Now factor the cubic polynomial \( x^3 + x^2 - 4x - 4 = 0 \). Try factoring by grouping:\[ x^2(x+1) - 4(x+1) = (x^2 - 4)(x+1) = (x-2)(x+2)(x+1) \].Set each factor to zero: \( x = 2, x = -2, x = -1 \).

Step 6: List All Real Solutions

Combine all zeros obtained:The zeros of the initial polynomial are \( x = 3, 2, -2, -1 \).

Key Concepts

Polynomial Equations

A polynomial equation involves a sum of terms, each consisting of a variable raised to a non-negative integer power, multiplied by a coefficient. In these equations, we aim to find the values of the variable that satisfy the equation, called the roots or solutions. For example, in the equation given:
\[x^4 - 2x^3 - 7x^2 + 8x + 12 = 0\],
we see a polynomial of degree 4. The degree of the polynomial indicates the highest power of the variable involved. Finding solutions to polynomial equations can often involve various techniques including the Rational Zero Theorem, synthetic division, and factoring.

Synthetic Division

Synthetic division is a simplified form of polynomial division, especially useful when dividing a polynomial by a binomial of the form \(x - c\). This method quickly checks potential roots identified by the Rational Zero Theorem.
To perform synthetic division:

• Write the coefficients of the polynomial.
• Place the potential root on the left side of the division line.
• Carry down the leading coefficient.
• Multiply this value by the potential root and add to the next coefficient.
• Continue this process across all coefficients.

A remainder of zero means the candidate is a root of the polynomial. This technique not only checks roots swiftly but also provides a reduced polynomial to continue solving for solutions.

Factoring Polynomials

Factoring polynomials involves writing the polynomial as a product of simpler polynomials that, when multiplied, give the original polynomial. This step is crucial after finding a zero using synthetic division, as it helps in reducing the equation to a simpler form.
For instance, after finding \(x = 3\) as a root, the polynomial reduces to \(x^3 + x^2 - 4x - 4\). Factoring can use common methods, such as:

• Factoring by grouping distinct terms to reveal a common factor.
• Using special identities, like the difference of squares.

In the example, group terms to factor:\[(x^2(x+1) - 4(x+1)) = (x-2)(x+2)(x+1) \,\].
Each factor indicates a solution, as when a factor equals zero, the entire expression will equal zero.

Real Solutions

Real solutions of a polynomial equation are the real numbers that satisfy the equation. After determining the factors of the reduced polynomial, set each factor to zero to find the solutions.
From the factors obtained by the example \((x - 2)(x + 2)(x + 1)\),
we set each equal to zero:

• \(x - 2 = 0 \rightarrow x = 2\)
• \(x + 2 = 0 \rightarrow x = -2\)
• \(x + 1 = 0 \rightarrow x = -1\)

Combining these with the previously found root \(x = 3\), the full set of real solutions for the polynomial equation are \(x = 3, 2, -2, -1\). Real solutions are essential in understanding the points where the polynomial graph intersects the \(x\)-axis.