Question

Solve each equation in the real number system. $$ 2 x^{4}-19 x^{3}+57 x^{2}-64 x+20=0 $$

Short answer

The roots are 5, 2, \( \frac{1}{2}, \) 2.

Step-by-step solution

Identify the Polynomial

Recognize that the given equation is a polynomial equation of the form \(2x^4 - 19x^3 + 57x^2 - 64x + 20 = 0\).

Find Possible Rational Roots

Using the Rational Root Theorem, identify possible rational roots. Possible rational roots are the factors of the constant term (20) divided by the factors of the leading coefficient (2). These roots could be \(\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20, \pm \frac{1}{2}, \pm \frac{5}{2}\).

Test the Possible Roots

Substitute each possible rational root into the polynomial to see if it equals zero. Start with \(x = 1, -1, 2, -2, \frac{1}{2},\) and so on.

Verify the First Root

If a root works, for example, if substituting \(x = 5\) gives zero, then \(x = 5\) is a root of the polynomial.

Factor the Polynomial

Use polynomial division or synthetic division to factor out \((x - 5)\). This will simplify the polynomial.

Repeat for Remaining Roots

Repeat steps 3-5 to find and verify other roots. If \(x = \frac{1}{2}\) is a root, then factor out \((2x - 1)\).

Solve Quadratic Equation

After factoring out known roots, solve the resulting quadratic equation to find the remaining roots.

Combine All Roots

Combine all the roots found. The roots for the polynomial \(2x^4 - 19x^3 + 57x^2 - 64x + 20 = 0\) are \(5, 2, \frac{1}{2}, 2\).

Key concepts

polynomial roots

The solution to a polynomial equation involves finding its roots. The roots of a polynomial are the values of \(x\) for which the polynomial equals zero. Sometimes, finding the roots requires testing various possibilities. For example, given the polynomial equation \(2x^4 - 19x^3 + 57x^2 - 64x + 20 = 0\), the task is to find the values of \(x\) that satisfy this equation.
Polynomial roots can come in various forms:

• Integer roots
• Fractional (rational) roots
• Irrational roots
• Complex roots

Identifying these roots allows us to break down the polynomial into simpler factors.

rational root theorem

The Rational Root Theorem is a key tool for discovering possible rational roots of a polynomial. According to this theorem, any rational root of a polynomial equation, given in standard form with integer coefficients, will be a fraction \(p/q\), where:

• \(p\) is a factor of the constant term (the last number in the polynomial)
• \(q\) is a factor of the leading coefficient (the first number in the polynomial)

For the polynomial \(2x^4 - 19x^3 + 57x^2 - 64x + 20 = 0\), the constant term is 20 and the leading coefficient is 2.
Possible rational roots are \(\frac{p}{q}\) values such as \(\frac{\text{factors of } \textbf{20}}{\text{factors of } \textbf{2}}\), resulting in possible roots of \(\text{\tiny\)\begin{pmatrix} \text{±1, ±2, ±4, ±5, ±10, ±20, ±\frac{1}{2}, ±\frac{5}{2}} \text{\tiny}\(\begin{pmatrix}}\).

synthetic division

Synthetic division is a simplified method of dividing a polynomial by a linear binomial of the form \((x - c)\). This method is efficient and less error-prone compared to long division. The steps are as follows:

• Write down the coefficients of the polynomial.
• Write the possible root outside the synthetic division bracket.
• Bring the leading coefficient down as is.
• Multiply this coefficient by the root and write the result under the next coefficient.
• Add these numbers and write the result below.
• Repeat the process until you reach the end of the polynomial.

If the remainder is zero, then the number outside (root) is a valid root. This helps to simplify the polynomial equation further. In the given example, once a root like \(x=5\) is found, the polynomial can be divided by \((x-5)\) to find the remaining polynomial of lower degree.

quadratic equations

After simplifying a higher-degree polynomial by factoring out linear terms, we often end up with quadratic equations. These are equations of the form \(ax^2 + bx + c = 0\). Quadratic equations can be solved using several methods:

• Factoring
• Completing the square
• Quadratic formula: \(x = \frac{-b \text{\tiny\)\begin{pmatrix} \text{±} \text{\tiny\(\begin{pmatrix} \text{±}\text{\tiny\)\begin{pmatrix}b^2-4ac}\text{\tiny})\textbf{}}}{\text{\tiny\(\begin{pmatrix}2a}\textbf{}}\).

Finding and learning how to solve quadratic equations is usually easier once the higher-degree polynomial has been factored down. In our example, after dividing by the found roots, we might be left with a quadratic polynomial, which can be solved for the remaining roots. The solutions (roots) of this quadratic equation added to previously found roots will give the complete solution to the original polynomial equation.