Problem 34 - Precalculus Mathematics for Calculus

Question

Find all rational zeros of the polynomial, and write the polynomial in factored form. $$P(x)=2 x^{3}-3 x^{2}-2 x+3$$

Step-by-Step Solution

Answer

Rational zeros are \( x = 1, \frac{3}{2}, -1 \); factored form is \( 2(x-1)(2x-3)(x+1) \).

Step 1: Apply the Rational Root Theorem

To find possible rational zeros, use the Rational Root Theorem. This theorem states that any rational zero of a polynomial \( a_n x^n + a_{n-1} x^{n-1} + \, ... \, + a_0 \) must be of the form \( \pm \frac{p}{q} \), where \( p \) is a factor of the constant term (\( a_0 = 3 \)) and \( q \) is a factor of the leading coefficient (\( a_n = 2 \)). Thus, the possible rational zeros are \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2} \).

Step 2: Test Possible Rational Zeros

Find which of these possible zeros are actual zeros by substituting them into \( P(x) = 2x^3 - 3x^2 - 2x + 3 \). Use direct substitution for the values \( x = 1, -1, \frac{1}{2}, -\frac{1}{2}, 3, -3, \frac{3}{2}, -\frac{3}{2} \).- For \( x=1 \), \( P(1) = 2(1)^3 - 3(1)^2 - 2(1) + 3 = 0 \). So, \( x = 1 \) is a zero.- Continue testing other values similarly.

Step 3: Perform Synthetic Division

With \( x = 1 \) as a zero, use synthetic division to divide \( P(x) \) by \( x - 1 \):1. Setup the synthetic division with coefficients: \( 2, -3, -2, 3 \).2. Process: Bring down the \( 2 \), multiply by \( 1 \), add to \( -3 \) to get \( -1 \), multiply by \( 1 \), add to \( -2 \) to get \( -3 \), multiply by \( 1 \), add to \( 3 \) to get remainder \( 0 \).The quotient is \( 2x^2 - x - 3 \).

Step 4: Factor the Quotient Polynomial

Find the zeros of the quadratic polynomial \( 2x^2 - x - 3 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -1, c = -3 \):\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \]\[ x = \frac{1 \pm \sqrt{1 + 24}}{4} \]\[ x = \frac{1 \pm \sqrt{25}}{4} \]\[ x = \frac{1 \pm 5}{4} \]Two solutions are found: \( x = \frac{6}{4} = \frac{3}{2} \) and \( x = \frac{-4}{4} = -1 \).

Step 5: Write the Polynomial in Factored Form

Now we have found all zeros: \( x = 1, \frac{3}{2}, -1 \). Therefore, the polynomial \( P(x) = 2x^3 - 3x^2 - 2x + 3 \) can be written in factored form as:\[ P(x) = 2(x - 1)\left(x - \frac{3}{2}\right)(x + 1) \]Adjust for the leading coefficient (2) embedded in the binomials to ensure the polynomial expands to its original form. Simplify to: \( 2(x - 1)(2x - 3)(x + 1) \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool for identifying potential rational zeros of a polynomial. It specifically applies to polynomials of the form \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0\). According to this theorem, any rational root \(\frac{p}{q}\) can be found by using the factors of the constant term of the polynomial, \(a_0\), for \(p\), and the factors of the leading coefficient, \(a_n\), for \(q\).

In our example where \(P(x) = 2x^3 - 3x^2 - 2x + 3\), the constant term \(a_0 = 3\) and the leading coefficient \(a_n = 2\).
By applying the Rational Root Theorem, we list possible rational roots as \(\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}\).

This list of possible roots helps narrow down the candidates that should be tested next in the polynomial equation.

synthetic division

Synthetic division provides an efficient method for dividing polynomials, particularly when dealing with polynomials of a higher degree. Once a potential root is found using the Rational Root Theorem, synthetic division can prove or disprove if the number is indeed a root.

For the polynomial \(P(x) = 2x^3 - 3x^2 - 2x + 3\), we use synthetic division to test \(x = 1\), which was found to be an actual root. Here's a basic walkthrough of the synthetic division process:

• Start by writing down the coefficients of the polynomial, \([2, -3, -2, 3]\).
• Use the potential root, \(x = 1\), placed on the left.
• Bring down the first coefficient (2), and then successively multiply the number you bring down by 1 and add to the next coefficient.
• The final number should be zero if it really is a root.

By applying these steps, you verify that \(x = 1\) is indeed a root, leaving you with a quotient of \(2x^2 - x - 3\), which is simpler to work with for further factorization.

factored form of polynomials

Writing a polynomial in its factored form is essential for understanding its roots. Once you have all the roots, you can express the polynomial as a product of factors.

For our example, with a polynomial \(P(x) = 2x^3 - 3x^2 - 2x + 3\), we found the roots \(x = 1, \frac{3}{2},\) and \( -1\).

• The factor corresponding to \(x = 1\) is \( (x - 1) \).
• The factor corresponding to \(x = \frac{3}{2}\) is \( (x - \frac{3}{2}) \).
• The factor corresponding to \(x = -1\) is \( (x + 1) \).

To ensure the leading coefficient is honored, each factor can be adjusted. In this case, incorporating the leading coefficient 2 results in each being present within a factor to quickly expand back to the original form. Hence, the complete factored version of \(P(x)\) is \(2(x - 1)(2x - 3)(x + 1)\).

This form is useful for quickly finding the roots and for understanding the behavior of the polynomial graph.