Question

Find all the real solutions of the equation. \(x^3-2 x^2-5 x+6=0\)

Short answer

The real solutions of the equation \(x^3 - 2x^2 - 5x + 6 = 0\) are '-1', '-2', and '3'.

Step-by-step solution

Guess the Root

A cubic equation always has at least one real root. In most high school math problems, there usually contains a root that is a small integer. An effective way to guess the potential root is to list the factors of the constant term. For this equation, the factors of '6' are +/-1, +/-2, +/-3, and +/-6. Check these numbers one by one by plugging them into the equation. Here, we can find that '-1' can make the equation hold, Hence '-1' is a root of the equation.

Synthetic Division

After guessing the root '-1', apply synthetic division to reduce the degree of the equation. Write down the coefficients of the original equation '1, -2, -5, 6' in a row, and put '-1' to the left. Then perform the synthetic division. The result of the division will give us a quadratic equation.

Solve the quadratic equation

After the synthetic division, a quadratic equation is obtained. Solve the quadratic equation to find the rest of the roots. In this case, the quadratic equation \(x^2 - x - 6 = 0 \) is gotten, which factors as \((x-3)(x+2) = 0\). So the solutions of this quadratic equation are '3' and '-2'

Determine All the Real Solutions

Therefore, by combining the real roots '-1' from the guess and '-2', '3' from the quadratic equation, the whole set of real solutions for the given equation are '-1', '-2', '3'.

Key concepts

Synthetic Division

Synthetic division is a shorthand method of dividing a polynomial by a binomial of the form \(x - c\). It simplifies the division process significantly by focusing on the coefficients of the polynomial. This technique is particularly useful when the divisor is straightforward, which often happens when we've already guessed one of the roots of a polynomial equation.

Here's how synthetic division works:

• Write down the coefficients of the polynomial. In our exercise, that's '1, -2, -5, 6'.
• Place the guessed root, '-1', to the left of these coefficients.
• The first coefficient, '1', is brought straight down.
• Multiply this value by the root, '-1', and write the result under the next coefficient, '-2'.
• Add the numbers to get the new second entry in the row.
• Repeat this process of multiply-and-add until you've completed the line.

This procedure ends with a remainder and a set of new coefficients that form a quadratic equation. In this example, the resulting quadratic from synthetic division is \(x^2 - x - 6\). The benefit of synthetic division is that it provides a much cleaner and quicker method to reduce the original cubic equation into a quadratic form, allowing further analysis and solution.

Factoring Quadratics

Once synthetic division provides a quadratic equation, the next step is to solve it. Factoring is the primary method used for these types of quadratic equations, particularly when they have rational roots.

For the quadratic \(x^2 - x - 6\), the task is to factor it into two binomials of the form \((x + m)(x + n)\).

The process typically involves:

• Identifying two numbers that multiply to the constant term ('-6') and add to the linear coefficient ('-1').
• After a bit of trial, you find that the numbers '-3' and '2' meet these criteria because \(-3 imes 2 = -6\) and \(-3 + 2 = -1\).
• Thus, the quadratic factors into \((x - 3)(x + 2)\), representing the roots of the quadratic equation.

Factoring quadratics is a crucial skill in algebra, as it transforms complex expressions into simpler ones. This step unlocks the additional real roots we found, '-3' and '2', critical to solving the original cubic equation.

Real Roots

When dealing with polynomial equations like the cubic equation \(x^3 - 2x^2 - 5x + 6 = 0\), finding real roots is one of the primary goals. A real root is a solution to the equation that is a real number, which means it doesn't have any imaginary part.

To find real roots effectively, you can:

• Check simple integer guesses based on the factors of the constant term.
• Apply methods like synthetic division to transform the equation.
• Solve the resulting quadratic equation by factoring or using the quadratic formula if necessary.

In this exercise, the real roots identified are '-1', '-2', and '3'. Each root satisfies the original equation, meaning when substituted back into the polynomial, the equation equals zero. Understanding how to find and verify real roots is essential not just for polynomial equations but for all algebraic calculations involving solutions.