Math Lesson 11.3.5 - How to Solve Polynomial Equations

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Welcome to our Math lesson on How to Solve Polynomial Equations, this is the fifth lesson of our suite of math lessons covering the topic of Solutions for Polynomial Equations, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Solving Polynomial Equations

So far, we have explained how to find all rational roots of a polynomial P(x). In fact, they represent nothing more but the solutions of the corresponding polynomial equation P(x) = 0. Hence, the above procedure is also used to solve polynomial equations of whatever degree. The only restriction is that all coefficients and the constant of the given polynomial must be integers. Otherwise, the equation must be solved using the computerized methods.

In short, all polynomial equations are solved by combining the Rational Root Theorem and the Synthetic Division Method, as explained earlier in this tutorial. Let's summarize everything through an example.

Example 2

1. Solve the following polynomial equation
   x³ - 2x² - 5x + 6 = 0
2. Write all possible factorisations for the left side of this equation.

Solution 2

1. Solving the above polynomial equation means finding all zeroes of the polynomial on the left side of the equation.
   The left part of this equation represents the third-degree polynomial
   P(x) = x³ - 2x² - 5x + 6
   where a_n = 1 and a₀ = 6.
   From the Rational Root Theorem, we know that all possible rational roots of a polynomial (if they exist) are in the form p/q, where p is a factor of a₀ and q is a factor of a_n.
   The possible factors of a₀ are ± 1, ± 2, ± 3 and ± 6, and those of a_n = ± 1. Thus, we obtain the following list of the possible zeroes of the polynomial:
   p/q = ± 1/± 1; ± 2/± 1; ± 3/± 1; ± 6/± 1
   After removing the duplicates, we can list eight possible roots for the polynomial P(x). They are -6, -3, -2, -1, 1, 2, 3 and 6.
   Now, let's see which of them is/are the true zeroes of the polynomial. Since the polynomial is of the third degree, we have three zeroes at maximum out of eight possible ones.
   Substituting each of them in the polynomial P(x) = x³ - 2x² - 5x + 6 yields
   P(-6) = (-6)³ - 2 ∙ (-6)² - 5 ∙ (-6) + 6
   = -216 - 2 ∙ 36 - 5 ∙ (-6) + 6
   = -216 - 72 + 30 + 6
   = -252
   P(-3) = (-3)³ - 2 ∙ (-3)²-5 ∙ (-3) + 6
   = -27 - 2 ∙ 9-5 ∙ (-3) + 6
   = -27 - 18 + 15 + 6
   = -24
   P(-2) = (-2)³-2 ∙ (-2)² - 5 ∙ (-2) + 6
   = -8 - 2 ∙ 4-5 ∙ (-2) + 6
   = -8 - 8 + 10 + 6
   = 0
   P(-1) = (-1)³ - 2 ∙ (-1)² - 5 ∙ (-1) + 6
   = -1 - 2 ∙ 1 - 5 ∙ (-1) + 6
   = -1 - 2 + 5 + 6
   = 10
   P(1) = (1)³ - 2 ∙ (1)² - 5 ∙ (1) + 6
   = 1 - 2 ∙ 1 - 5 ∙ 1 + 6
   = 1 - 2 - 5 + 6
   = 0
   P(2) = (2)³ - 2 ∙ (2)² - 5 ∙ (2) + 6
   = 8 - 2 ∙ 4 - 5 ∙ 2 + 6
   = 8 - 8 - 10 + 6
   = -4
   P(3) = (3)³ - 2 ∙ (3)² - 5 ∙ (3) + 6
   = 27 - 2 ∙ 9 - 5 ∙ 3 + 6
   = 27 - 18 - 15 + 6
   = 0
   P(6) = (6)³ - 2 ∙ (6)² - 5 ∙ (6) + 6
   = 216 - 2 ∙ 36 - 5 ∙ 6 + 6
   = 216 - 72 - 30 + 6
   = 120
   From the operations above it is clear that x = -2, x = 1 and x = 3 are the solutions (roots) of the given polynomial equation, as only for those we have P(x) = 0.
2. We have to use the Synthetic Division Method to find all possible factorisations of the polynomial on the left side of the equation. Obviously, we will consider only the already found roots for this purpose. Thus, using the coefficients 1, -2, -5 and 6 for the upper row and the roots of the equation on the left yields:
   1. For the root x = -2: The numbers in the last row (except the rightmost one that gives the remainder) represent the coefficients of the long expression in the brackets. The other expression involves the zero of the polynomial. Hence, we obtain
      P(x) = x³ - 2x² - 5x + 6 = (x + 2) ∙ (x² - 4x + 3)
   2. For the root x = 1: The numbers in the last row (except the rightmost one that gives the remainder) represent the coefficients of the long expression in the brackets. The other expression involves the zero of the polynomial. Hence, we obtain
      P(x) = x³ - 2x²-5x + 6 = (x - 1)(x² - x - 6)
   3. For the root x = 3: The numbers in the last row (except the rightmost one that gives the remainder) represent the coefficients of the long expression in the brackets. The other expression involves the zero of the polynomial. Hence, we obtain
      P(x) = x³ - 2x² - 5x + 6 = (x - 3)(x² + x - 2)

More Solutions for Polynomial Equations Lessons and Learning Resources

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4. Polynomials Revision Notes: Solutions for Polynomial Equations. Print the notes so you can revise the key points covered in the math tutorial for Solutions for Polynomial Equations
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6. Check your calculations for Polynomials questions with our excellent Polynomials calculators which contain full equations and calculations clearly displayed line by line. See the Polynomials Calculators by iCalculator™ below.
7. Continuing learning polynomials - read our next math tutorial: Rational Expressions

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