91Ó°ÊÓ

The Rational Root Theorem is a useful tool when solving polynomial equations, particularly when we want to find potential rational zeros of a polynomial. A rational root is a solution to the polynomial equation that can be expressed as a fraction \( \frac{p}{q} \), where both \(p\) and \(q\) are integers. This theorem provides a way to list out all possible rational roots: **any rational solution of the polynomial \( a_n x^n + a_{n-1} x^{n-1} + ... + a_0 = 0 \) is the ratio of a factor of the constant term \( a_0 \)** to a factor of the leading coefficient \( a_n \).

In the exercise provided, our polynomial is \( g(x) = 2x^3 - x^2 + 28x + 51 \). Here, the constant term \( a_0 \) is 51, and the leading coefficient \( a_n \) is 2. Thus, we take all divisors of 51 (\( \pm 1, \pm 3, \pm 17, \pm 51 \)) and divide them by all divisors of 2 (\( \pm 1, \pm 2 \)), yielding the possibilities: \( \pm 1, \pm 3, \pm 17, \pm 51, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{17}{2}, \pm \frac{51}{2} \).

By testing these values in the polynomial, one can quickly determine which, if any, are actual roots of the polynomial.

Synthetic division is a shortcut method for dividing a polynomial by a linear divisor of the form \( x - c \). It's less cumbersome than long division and is especially useful when you're verifying potential roots of a polynomial.

Here's how synthetic division works, using our example polynomial \( g(x) = 2x^3 - x^2 + 28x + 51 \) and the found root \( x = -3 \):

• Set up the synthetic division using the divisor \( x + 3 \) (as \( x = -3 \) means \( (x - (-3)) \)).
• Write the coefficients of the polynomial: 2, -1, 28, and 51.
• Bring down the leading coefficient (2) to the bottom row.
• Multiply this by -3 and add to the next coefficient (-1), repeating this cycle along the row.
• The final number you'll obtain is the remainder, and if it's zero, the division confirms \( x = -3 \) is indeed a root.

In this example, synthetic division produces a quotient of \( 2x^2 - 7x + 17 \) with a remainder of zero, proving \( x = -3 \) is a correct root.

Complex roots arise when we solve polynomial equations and find solutions involving the square root of a negative number. In the given polynomial exercise, after using synthetic division, we were left with a quadratic \( 2x^2 - 7x + 17 = 0 \).

To find the roots, we use the quadratic formula, but encounter a negative discriminant. The discriminant (the part under the square root in the quadratic formula) is calculated as \( b^2 - 4ac \). For \( 2x^2 - 7x + 17 \), substituting \( a = 2 \), \( b = -7 \), and \( c = 17 \) gives the discriminant as \( 49 - 136 = -87 \).

Because the discriminant is negative, our solutions will involve the imaginary unit \( i \), where \( i^2 = -1 \). Hence, the roots become complex numbers: \( x = \frac{7 \pm i\sqrt{87}}{4} \).

Complex roots in polynomials always appear in conjugate pairs when the coefficients are real, as seen in this example.

The quadratic formula is a fundamental tool for finding the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is expressed as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

We encounter the quadratic formula in this problem after obtaining a quadratic from synthetic division. Given the equation \( 2x^2 - 7x + 17 = 0 \), we substitute \( a = 2 \), \( b = -7 \), and \( c = 17 \) into the formula:

• Calculate \( -b \) which makes it \( 7 \).
• Compute the discriminant \( \sqrt{b^2 - 4ac} = \sqrt{49 - 136} = \sqrt{-87} \).
• Since the discriminant is negative, this implies the roots are complex.
• Finally, express the roots as \( x = \frac{7 \pm i\sqrt{87}}{4} \).

The quadratic formula reveals both real and complex roots, providing us with comprehensive solutions to quadratic equations.