Polynomial equations

Polynomial equations are equations that can be written in the form \( P(x)=0 \), where \( P(x) \) is a polynomial in the variable x with numerical coefficients. In this section, we focus on the simplest case: first-degree polynomial equations in one variable.

In plain terms, the equations discussed here satisfy two basic conditions:

• An equation is "polynomial" when the variable appears only with non-negative integer powers and never in the denominator.

  Note. Coefficients may be integers or fractions. What matters is that the variable itself does not appear in denominators or under radicals.

• The equation has numerical coefficients, meaning that no literal parameters are involved.

Understanding with an example

Let’s start with a simple polynomial equation of first degree:

$$ \frac{1}{2} x + 3 = \frac{15}{2} $$

This is a polynomial equation because the variable x appears only to the first power and never in the denominator.

It is also a first-degree equation, since the highest power of x is 1.

Example 2

Now look at this equation:

$$ \frac{1}{2x} + 3 = \frac{15}{2} $$

Here, the variable x appears in the denominator, so the equation is not polynomial.

Therefore, it is not a polynomial equation.

Example 3

Consider the following equation:

$$ \frac{a}{2} x + 3 = \frac{15}{2} $$

This equation is polynomial in x, but it is not numerical because the coefficient of x contains a literal parameter (a).

So, in this context, it is excluded from the class under discussion.

First-degree polynomial equations

A first-degree polynomial equation in one variable is an equation in which the highest power of x is 1. A typical example is $$ 2x = 8 $$

Any first-degree polynomial equation can be reduced to the standard form ax = b:

$$ ax = b $$

Here, “a” is the coefficient of x and “b” is the constant term.

How to solve them

If the coefficient “a” is not zero (a≠0), the equation is called determinate because it has exactly one solution:

$$ x = \frac{b}{a} $$

Note. To isolate x, we apply the second principle of equivalence and divide both sides by “a”: $$ ax = b $$ $$ \frac{ax}{a} = \frac{b}{a} $$ Simplifying: $$ \require{cancel} \frac{\cancel{a}x}{\cancel{a}} = \frac{b}{a} $$ $$ x = \frac{b}{a} $$

Special cases

If the coefficient “a” is zero (a=0), two distinct situations may arise:

• Indeterminate equation: infinitely many solutions, when b is also zero (b=0).

Example. $$ 0x = 0 $$ For any value of x, both sides remain equal to zero. The equation is therefore true for all x.

• Impossible equation: no solutions, when b is not zero (b≠0).

Example. $$ 0x = 2 $$ The left-hand side is always zero, while the right-hand side is not. The equation has no solutions.

Worked example

Let’s solve the following first-degree polynomial equation:

$$ 5x - 1 = 2x + 5 $$

We apply the first principle of equivalence and subtract 2x from both sides:

$$ 5x - 1 - 2x = 2x + 5 - 2x $$

$$ 3x - 1 = 5 $$

Then we add 1 to both sides:

$$ 3x - 1 + 1 = 5 + 1 $$

$$ 3x = 6 $$

Now the equation is in the standard form ax = b.

Since both the coefficient and the constant term are nonzero, the equation is determinate and has exactly one solution.

Dividing both sides by 3:

$$ \frac{3x}{3} = \frac{6}{3} $$

$$ \frac{\cancel{3}x}{\cancel{3}} = 2 $$

$$ x = 2 $$

So, the solution is x = 2.

Check. Substitute x=2 into the original equation: $$ 5x - 1 = 2x + 5 $$ $$ 5 \cdot 2 - 1 = 2 \cdot 2 + 5 $$ $$ 9 = 9 $$ The equation is satisfied, confirming that x=2 is the correct solution.