Introduction to Polynomials
A polynomial in one variable \(x\) of degree \(n\) is of the form:
\(f(x) = {a_0}{x^n} + {a_1}{x^{n  1}} +{a_2}{x^{n  2}} + ... + {a_{n  1}}x + {a_n}\)
where \(a_0,a_1,.....a_{n1},a_n\) are coefficient of the \(n\) terms of the polynomial and \(a_0 \ne 0\)
Zeros of Polynomial
The zeros of a polynomial are those values of the variable for which the polynomial as a whole has zero value.
For example, consider the polynomial:
\(p\left( x \right) = x+2 \)
The zero of the polynomial is \(2\) because when \(x\) = \(2\),
\(p\left( x \right) = 0\)
Consider another example:
\(p\left( x \right) = x^2  9 \)
The zeros of the polynomial are \(\pm 3\)
When \(x = \pm 3\)
\(p\left( x \right) =3^2  9 = 0 \)
Note that when the coefficient of all the terms and the constant of the polynomial is \(0\), the polynomial is called a zero polynomial and is denoted by 0
Quadratic Polynomial Formula
A quadratic polynomial is of the form \[\begin{align}&p\left( x \right):a{x^2} + bx + c \end{align}\] where \(a \neq 0 \)
Note: "Quad" means 2
The degree of a polynomial has to be 2
Hence, \(a\) which is the coefficient of \(x^2\) cannot be 0
The quadratic formula to find the solution to a quadratic equation is given by:
\(\begin{align}x =\frac{b \pm \sqrt{b^24ac}}{2a}\end{align}\) 
 \(a\) is the coefficient of \(x^2\) and is also called as the quadratic coefficient
 \(b\) is the coefficient of \(x\) and is also called as the linear coefficient and
 \(c\) is the constant term
Sum of Roots of a Quadratic Polynomial
If \(\alpha\) and \(\beta \) are the zeros of the quadratic polynomial \[\begin{align}&f\left( x \right):a{x^2} + bx + c \end{align}\]
The sum of the roots of the polynomial is:
\( \begin{align} \alpha +\beta = \frac{b}{a}\end{align}\) 
In other words, it refers to: \[ \dfrac{\text{coefficient of }x}{\text{coefficient of }x^2}\]
Product of Roots of a Quadratic Polynomial
If \(\alpha\) and \(\beta \) are the zeros of the quadratic polynomial \[\begin{align}&f\left( x \right):a{x^2} + bx + c \end{align}\]
The product of the roots of the polynomial is:
\( \begin{align}\alpha \: \beta = \frac{c}{a}\end{align}\) 
In other words, it refers to: \[\frac{\text{constant term}}{\text{coefficient of }x^2}\]
Relation between the Coefficient and the Sum and Product of the Zeros
Now that we have seen the crucial role played by the coefficients in determining the characteristics of any polynomial, we turn our attention to a special case: quadratic polynomials.
In particular, we want to see how the zeroes of a quadratic polynomial are related to its coefficients.
To understand that relation between the coefficient and the zeros, let’s consider some examples.
Let,
\[a\left( x \right): {x^2}  3x + 2\]
The coefficient of x in this polynomial is –3, while the constant term is 2
To determine the zeroes of this polynomial, we factorize it:
\[\begin{array}{l}a\left( x \right) = \left( {x  1} \right)\left( {x  2} \right)\\{\rm{Zeroes}}:\;x = 1,\;x = 2\end{array}\]
Note that the sum of the zeroes is 3, which is the negative of the coefficient of x in the polynomial, while the product of the zeroes is 2, the same as the constant term.
Is there some relation here?
Let’s take another example.
Consider
\[b\left( x \right): {x^2}  7x + 6\]
The zeroes can be determined by factorizing:
\[\begin{array}{l}b\left( x \right) = \left( {x  1} \right)\left( {x  6} \right)\\{\rm{Zeroes}}:\;x = 1,\;x = 6\end{array}\]
The sum of the zeroes is 7, once again the negative of the coefficient of x, while their product is 6, equal to the constant term of the polynomial.
This cannot be a coincidence, so let us explore this more formally.
Consider the following quadratic polynomial:
\[p\left( x \right): \left( {x  a} \right)\left( {x  b} \right)\]
The zeroes are a and b.
Let us denote the sum and product of the zeroes by s and p respectively:
\[\begin{array}{l}S = a + b\\P = ab\end{array}\]
Next, we note that the polynomial can be written (by multiplication) as follows:
\[\begin{array}{l}p\left( x \right) = {x^2}  ax  bx + ab\\ \Rightarrow \;\;\;p\left( x \right) = {x^2}  \left( {a + b} \right)x + ab\\ \Rightarrow \;\;\;p\left( x \right) = {x^2}  Sx + P\end{array}\]
Now the relation should be clear!
The coefficient of x in the polynomial is the negative of the sum of its roots, while the constant term is the same as the product of the roots.
Note that we assumed the polynomial p to be of the form
\[p\left( x \right): \left( {x  a} \right)\left( {x  b} \right)\]
That is, the coefficient of the square term in this polynomial is 1
However, in an arbitrary quadratic polynomial, this coefficient can be any nonzero real number.
Therefore, let us generalize our result.
Consider the following quadratic polynomial:
\[p\left( x \right): l{x^2} + mx + n\]
The coefficient of the square term is l.
Let the zeroes of this polynomial be a and b.
Then, we can also write it as follows (note this very carefully!):
\[p\left( x \right): l\left( {x  a} \right)\left( {x  b} \right)\]
Make sure you understand this.
Now, we compare the two expressions (corresponding to the same polynomial):
\[\begin{align}&l{x^2} + mx + n = l\left( {x  a} \right)\left( {x  b} \right)\\&\Rightarrow \;\;\;{x^2} + \frac{m}{l}x + \frac{n}{l} = \left( {x  a} \right)\left( {x  b} \right)\\&\Rightarrow \;\;\;{x^2} + \frac{m}{l}x + \frac{n}{l} = {x^2}  \left( {a + b} \right)x + ab\\&\Rightarrow \;\;\;{x^2} + \frac{m}{l}x + \frac{n}{l} = {x^2}  Sx + P\\&\Rightarrow \;\;\;  S = \frac{m}{l},\;P = \frac{n}{l}\\&\Rightarrow \;\;\;\left\{ \begin{gathered}S =  \frac{m}{l} =  \frac{{{\rm{coeff}}\;{\rm{of}}\;x}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}}\\\!\!\!\!\!\!\!\!\!\!P = \frac{n}{l} = \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}}\end{gathered} \right.\end{align}\]
This is an extremely important result.
In any quadratic polynomial:

The sum of the zeroes is equal to the negative of the coefficient of x by the coefficient of x^{2}.

The product of the zeroes is equal to the constant term by the coefficient of x^{2}.
Knowing how we derive this result is crucial, so make sure that you fully understand it.
Note a particular case.
If the coefficient of \({x^2}\) is 1, that is, if the polynomial is of the following form:
\[p\left( x \right): {x^2} + ex + f,\]
then the sum and product of the zeroes are simply:
\[\begin{align}&S =  \frac{e}{1} =  e\\&P = \frac{f}{1} = f\end{align}\]
Solved Examples of Quadratic Polynomials
Example 1 
Consider the following quadratic polynomial:
\[p\left( x \right): {x^2}  13x + 42\]
Determine the sum and product of the zeroes using the coefficients, and then verify your answer by factorization.
Solution:
We have:
\[\begin{align}&S =  \frac{{{\rm{coeff}}\;{\rm{of}}\;x}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} =  \frac{{\left( {  13} \right)}}{1} = 13\\&P = \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} = \frac{{42}}{1} = 42\end{align}\]
Now, we factorize the polynomial by splitting the middle term:
\[\begin{align}&p\left( x \right) = {x^2}  13x + 42\\& = {x^2}  6x  7x + 42\\& = x\left( {x  6} \right)  7\left( {x  6} \right)\\& = \left( {x  6} \right)\left( {x  7} \right)\end{align}\]
The zeroes are 6 and 7
\(\therefore\) Sum of the zeros is 13 and their product is 42 
Example 2 
Consider the following quadratic polynomial:
\[q\left( x \right): 2{x^2} + 5x + 2\]
Determine the sum and product of the zeroes using the coefficients, and then verify your answer by factorization.
Solution:
We have:
\[\begin{align}&S =  \frac{{{\rm{coeff}}\;{\rm{of}}\;x}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} =  \frac{5}{2}\\&P = \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} = \frac{2}{2} = 1\end{align}\]
Now, we factorize the polynomial by splitting the middle term:
\[\begin{array}{l}q\left( x \right) = 2{x^2} + 5x + 2\\\;\;\;\;\;\,\,\, = 2{x^2} + 4x + x + 2\\\,\,\,\,\,\,\,\;\,\, = 2x\left( {x + 2} \right) + x + 2\\\,\,\,\,\,\,\,\;\,\, = \left( {2x + 1} \right)\left( {x + 2} \right)\end{array}\]
The zeroes are:
\[x =  \frac{1}{2},\;\;x =  2\]
Their sum and product is clearly the same as the values we obtained using the coefficients:
\[\begin{array}{l}S = \left( {  \frac{1}{2}} \right) + \left( {  2} \right) =  \frac{5}{2}\\P = \left( {  \frac{1}{2}} \right) \times \left( {  2} \right) = 1\end{array}\]
\(\therefore\) Sum = \(\dfrac{5}{2}\) and Product = 1 
Example 3 
Let the sum and product of zeroes of a quadratic polynomial be a and b.
Determine this polynomial.
Solution:
Recall that if the coefficient of the square term in a quadratic polynomial is unity, then the sum of the zeroes is simply the negative of the coefficient of x, while the product of the zeroes is simply equal to the constant term.
Keeping this observation in mind, consider the following polynomial:
\[p\left( x \right): {x^2}  ax + b\]
Clearly, the sum of the zeroes of this polynomial will be a, while their product will be b.
Thus, this polynomial is an answer to our question.
But this is only one such polynomial, in which we took the coefficient of the square term to be 1.
To obtain the general answer, simply multiply this polynomial by any nonzero real number k:
\[\begin{array}{l}p\left( x \right): k\left( {{x^2}  ax + b} \right)\\ \Rightarrow \;\;\;\;p\left( x \right):\;\;\;k{x^2}  kax + kb\end{array}\]
This is the complete answer to our question. Verification:
\[\begin{align}&S =  \frac{{{\rm{coeff}}\;{\rm{of}}\;x}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} =  \frac{a}{1} = a\\&P = \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} = \frac{b}{1} = b\end{align}\]
\(\therefore\) \(p\left( x \right): k\left( {{x^2}  ax + b} \right) \) 
Example 4 
Determine a quadratic polynomial, the sum and product of whose zeroes are 15 and 56 respectively.
Solution:
Based on the solution to the problem above, we can simply write the answer as:
\[p\left( x \right): k\left( {{x^2}  15x + 56} \right)\]
Note that we could have determined k if we had some more information, as in the problem that follows.
\(\therefore\) \(p\left( x \right): k\left( {{x^2}  15x + 56} \right)\) 
Example 5 
The sum and product of the zeroes of a quadratic polynomial p are 9 and 20 respectively.
Also,\(p\left( 6 \right) = 4\).
Determine the polynomial \(p\left( x \right)\).
Solution:
Note that in this problem, we have been given some extra information – the value of the polynomial at a particular x value.
Using the sum and product values, we can write the polynomial as:
\[p\left( x \right): k\left( {{x^2}  9x + 20} \right)\]
Now,
\[\begin{array}{l}p\left( 6 \right) = 4\\ \Rightarrow \;\;\;k\left( {{6^2}  9\left( 6 \right) + 20} \right) = 4\\ \Rightarrow \;\;\;k\left( {36  54 + 20} \right) = 4\\ \Rightarrow \;\;\;2k = 4\\ \Rightarrow \;\;\;k = 2\end{array}\]
Note how the extra information enabled us to determine the value of k.
The polynomial therefore is:
\[\begin{array}{l}p\left( x \right): 2\left( {{x^2}  9x + 20} \right)\\ \Rightarrow \;\;\;p\left( x \right):\;\;\;2{x^2}  18x + 40\end{array}\]
\(\therefore\) \(p\left( x \right) = {2{x^2}  18x + 40} \) 
Example 6 
Consider the following quadratic polynomial:
\[p\left( x \right): {x^2} + x + 1\]
What is the sum and product of the zeroes of this polynomial?
Solution:
We have:
\[\begin{align}& S =  \frac{{{\text{coeff}}\;{\text{of}}\;x}}{{{\text{coeff}}\;{\text{of}}\;{x^2}}} =  \frac{1}{1} =  1 \\& P = \frac{{{\text{constant}}}}{{{\text{coeff}}\;{\text{of}}\;{x^2}}} = \frac{1}{1} = 1 \hfill \\ \end{align} \]
Simple enough, but we have a problem.
This polynomial does not have real zeroes!
We have discussed earlier how some quadratic polynomials may not have real zeroes (they do have zeroes, but these zeroes happen to be nonreal, that is, complex).
In this case, if we apply the quadratic formula (this has also been discussed earlier, and will be discussed in much more detail in a subsequent chapter), we see that the zeroes of this polynomial will be:
\[\begin{align}&a = 1,b = 1,\;c = 1\\& \Rightarrow \;\;\;x = \frac{{  1 \pm \sqrt {{1^2}  4 \times 1 \times 1} }}{{2 \times 1}}\\&= \frac{{  1 \pm \sqrt {  3} }}{2}\end{align}\]
Observe that we have a negative quantity (–3) under the squareroot sign.
Thus, the zeroes are nonreal.
This does not mean that the zeroes are invalid, or do not exist, or something like that.
This simply means that the zeroes are not real numbers, but they are still perfectly valid zeroes.
However, we have seen the sum and product of the zeroes of this polynomial to be real numbers (above).
So what is happening here?
The zeroes are nonreal, but their sum and product seem to be real.
Are we going wrong somewhere?
The answer is: no.
What we have done above in calculating the sum and product of the zeroes is perfectly correct.
The zeroes are nonreal, but even then, their sum and product will be real.
This should not be surprising.
For example, if two irrational numbers sum to a rational number, you would not have been surprised.
Similarly, there should be no surprise if two nonreal numbers add to a real number.
To summarize, the relation between the sum and product of zeroes, and the coefficients of the polynomial, is universally true – it works in all cases, even if the zeroes themselves are nonreal.
\(\therefore\) Sum = 1 and Product = 1 and Imaginary roots. 
Example 7 
Consider the following polynomial:
\[p\left( x \right): {x^2}  2x + 3\]
What is the sum of the squares of the zeroes of this polynomial?
Solution:
To solve this problem, you might think that we need to calculate the two zeroes separately, and then square and add them to obtain our answer.
However, the zeroes of this polynomial are nonreal, as can be verified easily.
Therefore, we take a different approach.
Let the two (nonreal) zeroes be α and β.
We have:
\[\begin{align}&\alpha + \beta =  \frac{{{\rm{coeff}}\;{\rm{of}}\;x}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} = 2\\&\alpha \beta = \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} = 3\end{align}\]
Now, we have:
\[\begin{align}&{\left( {\alpha + \beta } \right)^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta \\& \Rightarrow \;\;\;{\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2}  2\alpha \beta \\& \Rightarrow \;\;\;{\alpha ^2} + {\beta ^2} = {2^2}  2 \times 3\\&= 4  6\\ & =  2\end{align}\]
Do not be surprised at this (negative) answer!
This is because the zeroes were nonreal, so there is no reason why their squares should add up to a positive number.
\(\therefore\) Sum of the squares of the zeros = \(2\) 
 The graph shows the xintercept of a quadratic equation and a cubic equation.
Find the sum of the roots of the cubic equation.
Also find the product of the roots of the polynomial represented in the graph (a) and graph (b)
Practice Questions
Here are a few activities for you to practice.
Select/Type your answer and click the "Check Answer" button to see the result.

The quadratic formula to find the roots of a quadratic equation \(\begin{align}p\left( x \right):a{x^2} + bx + c \end{align}\) is \(\begin{align}{b \pm \sqrt{b^24ac} \over 2a}\end{align}\)

The sum of the zeroes of a quadratic polynomial is equal to the negative of the coefficient of \(x\) by the coefficient of \(x^2\).

The product of the zeroes is equal to the constant term by the coefficient of \(x^2\)

A polynomial having value \(0\) is called a zero polynomial.
Frequently Asked Questions(FAQs)
1. What is the sum of the roots of a cubic equation?
A cubic equation is of the form \(ax^3 +bx^2 +cx +d = 0\)
The sum of the roots of the cubic equation is \(\begin{align}\frac{b}{a}\end{align}\)
That is: \( \begin{align}\frac{\text{coefficient of }x^2}{\text{coefficient of } x^3}\end{align}\)
2. How do you find the zeros of a polynomial?
Step 1: Use the Rational Zero Theorem to list all possible rational zeros of the polynomial.
Rational zero of the polynomial =
\((\frac{\text{constent term of the polynomial}}{\text{leading coefficient of the polynomial}})\)
Step 2: Use synthetic division to evaluate all given possible zeros
Step 3: Repeat step 2 using the quotient found with synthetic division; continue until the quotient is a quadratic polynomial.
step 4: Use the quadratic formula to find the factors of the quadratic polynomial
3. What are the zeroes of a polynomial?
The zeroes of a polynomial are those values of the variable for which the polynomial as a whole has zero value.
Example: Consider the polynomial:
\(p\left( x \right) = x^2  16 \)
The zeros of the polynomial are \(\pm 4\)
When \(x = \pm 4\)
\(p\left( x \right) = 0 \)
4. How many zeros does a polynomial have?
A polynomial of \(n\) degree has \(n\) zeros.
 A linear polynomial has 1 zero.
 A quadratic polynomial has 2 zeros.
 A cubic polynomial has 3 zeros.
 A polynomial of degree 8 has 8 zeros.