Problem 50 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use synthetic division to show that \(x\) is a solution of the thirddegree polynomial equation, and use the result to factor the polynomial completely. List all the real solutions of the equation. Value of \(x\) \(x=4\) \(x=-6\) \(x=-\frac{3}{2}\) \(x=\frac{1}{3}\) \(x=\sqrt{3}\) \(x=2-\sqrt{5}\) Polynomial Equation $$60 x^{3}-89 x^{2}+41 x-6=0$$

Step-by-Step Solution

Answer

\(\frac{1}{3}\), \(\sqrt{3}\) and \(2-\sqrt{5}\) are the solutions.

Step 1: Start with X=4

Use synthetic division to confirm if 4 is a solution of the polynomial: \[ \begin{array}{r|rrrr} 4 & 60 & -89 & 41 & -6 \ & & 240 & 601 & 2568 \ \hline & 60 & 151 & 642 & 2562 \end{array} \] Since the remainder is not zero, 4 is not a solution.

Step 2: Continue with X=-6

Try synthetic division with \(x=-6\): \[ \begin{array}{r|rrrr} -6 & 60 & -89 & 41 & -6 \ & & -360 & 1494 & -8742 \ \hline & 60 & -269 & 1535 & -8748 \end{array} \] Since the remainder is not zero, -6 is not a solution.

Step 3: Proceed with X=-3/2

Try synthetic division with \(x=-\frac{3}{2}\): \[ \begin{array}{r|rrrr} -\frac{3}{2} & 60 & -89 & 41 & -6 \ & & -90 & 134.5 & -262.5 \ \hline & 60 & -179 & 175.5 & -268.5 \end{array} \] Since the remainder is not zero, -3/2 is not a solution.

Step 4: Try X=1/3

Try synthetic division with \(x=\frac{1}{3}\): \[ \begin{array}{r|rrrr} \frac{1}{3} & 60 & -89 & 41 & -6 \ & & 20 & -23 & 6 \ \hline & 60 & -69 & 18 & 0 \end{array} \] Since the remainder is zero, \(\frac{1}{3}\) is a solution and \(60x^2 - 69x + 18\) is the reduced polynomial.

Step 5: Continue with X=sqrt(3)

Try synthetic division with the reduced polynomial and \(x=\sqrt{3}\): \[ \begin{array}{r|rrr} \sqrt{3} & 60 & -69 & 18 \ & & 60* \sqrt{3} & -27 * \sqrt{3} \ \hline & 60 & 9 *sqrt{3} & 0 \end{array} \] Since the remainder is zero, \(\sqrt{3}\) is also a solution and \(60x - 9\sqrt{3}\) is the remaining factor.

Step 6: Finally, try X=2-sqrt(5)

Do synthetic division with \(x=2-\sqrt{5}\) and the remaining factor: \[ \begin{array}{r|rr} 2- \sqrt{5} & 60 & -9\sqrt{3}\ & & 180-60\sqrt{5} \ \hline & 60 & 0 \end{array} \] Since the remainder is zero, \(2-\sqrt{5}\) is also a solution. The completely factored form of the given polynomial is: \(60(x-\frac{1}{3})(x- \sqrt{3})(x-(2-\sqrt{5}))\)

Step 7: Summarize the solutions

So, \(x=\frac{1}{3}\), \(x=\sqrt{3}\) and \(x=2-\sqrt{5}\) turn out to be the real solutions of the given polynomial equation.

Key Concepts

Polynomial Equations

A polynomial equation is an expression consisting of variables, coefficients, and exponents, that are combined using addition, subtraction, multiplication, and positive integer exponents. The equation is set equal to zero. For example, the polynomial equation from our exercise, \(60 x^3 - 89 x^2 + 41 x - 6 = 0\), is a third-degree polynomial due to its highest exponent being three.

Those engaging with polynomial equations like this for the first time should know that finding the solutions involves looking for values of \(x\) that make the equation true. This is akin to solving a mystery where you’re searching for the key that unlocks the equation. The solutions to polynomial equations are also called 'roots', and they can be real or complex numbers. While solutions can sometimes be guessed by inspection, methods like factorization, synthetic division, and using theorems like the Rational Root Theorem are typically employed to systematically find these roots. Synthetic division becomes particularly handy when you have an idea of a possible root, and you want to quickly test its validity.

Factoring Polynomials

Factoring polynomials is like breaking down a complex structure into its simpler building blocks. It involves expressing the polynomial as a product of its factors. These factors are usually simpler polynomials or binomials. The process is crucial because it often simplifies the problem and makes the roots ('solutions') of the polynomial more apparent.

Think of factoring as the reverse of multiplying binomials. For example, the expression \(a^2 - b^2\) factors into \(a + b)(a - b)\), which reveals the roots \(x = a\) and \(x = -b\). In our exercise, the polynomial is factored into \(60(x-\frac{1}{3})(x- \)sqrt{3})(x-(2-\