Sum and Product of Zeroes in a Quadratic Polynomial
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, 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:
\[\begin{align}&S =  \frac{e}{1} =  e\\&P = \frac{f}{1} = f\end{align}\]
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. Their sum 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:
\[\begin{array}{l}S = \left( {  \frac{1}{2}} \right) + \left( {  2} \right) =  \frac{5}{2}\\P = \left( {  \frac{1}{2}} \right) \times \left( {  2} \right) = 1\end{array}\]
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 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:
\[\begin{align}&S =  \frac{{{\rm{coeff}}\;{\rm{of}}\;x}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} =  \frac{1}{1} =  1\\&P = \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}} = \frac{1}{1} = 1\end{align}\]
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.
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:
\[\begin{array}{l}p\left( x \right): 2\left( {{x^2}  9x + 20} \right)\\ \Rightarrow \;\;\;p\left( x \right):\;\;\;2{x^2}  18x + 40\end{array}\]
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 earlier discussed 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.
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.