A polynomial equation is an equation that can be written as an equation with 0 on one side and a polynomial on the other, for example 2a + 5 = 0.
Polynomials are distinguished in two ways, by the number of terms and its degree.
Consider the polynomial 2x + 3x + 4. It has 3 separate monomial terms: 2x, 3x, 4 so it is a polynomial of three
terms or a trinomial. The degree of a polynomial is the highest degree of any of its monomial terms. The degree of a
monomial is the sum of the exponents of its variables. In our example we have:
2x, which has 1 variable of degree 1, so its degree is 1
3x, which has 1 variable of degree 1, so its degree is 1
4, which has 0 variables since it is a constant, so its degree is 0
The highest degree of the monomials is 1, so the polynomial 2x + 3x + 4 is of degree 1 and is a trinomial.
A polynomial with 2 terms is also called a binomial. None of the other polynomials have special names based on
the number of terms they have.
Polynomial equations also have special names based on the degree of the equation. Let us determine the degree of
a few more example equations:
4x + 5 = 0
4x is the highest degree monomial term with a degree of 1, so the polynomial equations degree is 1.
3x + 2xy = 0
2xy is the highest degree monomial term with a degree of 2, since it has 2 variables each with a power of 1, so the
degree of this polynomial equation is 2.
2x2 + 3x + 4 = 0
2x2 is the highest degree monomial term with a degree of 2, since x has an exponent of 2, so the degree of this
polynomial equation is 2.
2x2y3 + 4x + 8 = 0 2x2y3 is the highest degree monomial term with a degree of 5 since the exponents of x and y
add up to 5, so the degree of this polynomial equation is 5.
We have already looked at linear equations, i.e., polynomial equations of degree 1; now let us look at some
polynomial equations of degree 2.
Sometimes f(x) may be a quadratic function of the form ax2+bx+c=0. This way of writing a quadratic equation with
highest to lowest degree from left to right is called standard form.
1.8.1. Factoring a quadratic equation to solve for the values of x
To solve these types of quadratic equations, we need to find 2 numbers p and q such that p+q=b and pq =ac.
Example: Find the solution of the quadratic equation -1=2t2 +3t
Answer
- Writing in standard form, 2t2 +3t +1=0
- Two numbers p and q so that pq=2 and p+q=3 are 2 and 1.
- Splitting the middle term 3t as 2t +t, we get 2t2 +2t+t +1=0
- We factor the quadratic equation so that it is in the form of (kt + r)(st +u) where k, r, s, u are any real numbers.
To factor 2t2 +2t+t +1 =0, we group the first two and the last two terms.
2t (t+1) +(t+1)=0
Taking (t+1) as common, (t+1)(2t+1)=0
Either (t+1) = 0 or (2t+1) = 0, so the solutions to the quadratic equation are t = -1 or t = -1/2.
1.8.2. Interpreting the solution of a quadratic equation
When we plot the graph of f(x) = ax2+bx+c for different values of x, we will get a curve. The points on the curve
represent the value of f(x) for a given value of x.
When we solve a quadratic equation to find its roots, we actually find the values of x for which f(x) =0. The values of x
obtained when f(x) = 0 are the x intercepts of the graph of f(x) i.e. they are the values of x where the graph cuts the
x axis. Suppose the roots of the equation f(x) =0 are x1 and x2 then it means f(x1) = f(x2) = 0. This means (x1, 0) and
(x2, 0) are points where the graph of f(x) meets the x axis.
Hence these roots of the quadratic equation are also called x intercepts.
Example
The function s (t) = t2 -2t gives the relation between distance 's' meters covered by a vehicle in time 't' seconds. Find
how much time is required to cover a distance of 24 m.
Answer
Substituting s as 24:
24= t2 -2t
Writing the equation in standard form i.e. from highest degree to lowest degree, t2 -2t-24=0
To find the factors of this quadratic equation, we need to find two numbers say p and q so that p+q=-2 and pq=-24
Writing factors of -24, the two factors that satisfy the above condition are -6 and 4.
The given equation can be written as t2 -6t+4t-24=0
Combining first two and last two terms: t (t-6)+4(t-6)=0
(t - 6)(t+4)=0
Therefore either t – 6 = 0, or t + 4 = 0, since if two numbers multiply together to equal zero then one of them must be
equal to zero
t≠ -4 since time cannot be measured as a negative quantity, hence t=6 seconds. This means that vehicle can travel
24 meters in 6 seconds.
1.8.3 Useful Identities
There are a few identities which are easy to recognize and can be factored easily. You should memorize these.
- x2+a2 +2ax=0
This type of equation can always be factored as (x + a) (x + a)=0 or (x + a)2 =0
- Similarly if the quadratic equation is of the form x2+a2 -2ax=0, it can be written as (x-a)2
- x2 - a2 = 0
Since x=a or –a, it can be factored as (x + a) (x-a) = 0
1.8.4. Completing the square
Sometimes it is not possible to factor a quadratic equation. One way to solve it is to transform it into the form (x ± a)2 = k
Then it is easy to find the value of x by first finding the square root of both the sides and then solving the linear equation.
Let the given quadratic equation be x2+ b +cx = 0. To convert it into the form of x2+ a2 +2ax= 0 we compare the
coefficient of x and find the value of a. Note that the coefficient of x2 must be 1, so if it is not in the equation that you
are given, divide the equation by whatever the coefficient of x2 is first in order to start with the appropriate form.
Then comparing cx with 2ax, we get a=c/2
Example 9
Find the x intercepts of the graph of the function f(x) = x2-6x+7.
Solution
To find the x intercepts, we substitute f(x)= 0.
x2 -6x + 7 cannot be factored as 7 cannot be written as the product of two numbers whose sum is -6. We use the
steps for completing the square mentioned above.
- Comparing -6x with 2ax, a=-3
- Adding a2 or 9 on both the sides, x2-6x+7+9=9 …………..Equation 1
- Subtracting 7 from both the sides, x2-6x+9=9-7=2……….Equation 2
- Comparing left side with x2+a2 -2ax, we can write the Equation 2 as (x-3)2=2…Equation 3
- To find the value of x, take the square root of both sides of equation 3
