Sum and Product of Zeroes in a Quadratic Polynomial
There are always two zeros for any quadratic polynomial. The zeros of a polynomial are those values of the variable for which the polynomial as a whole has zero value. The sum and product of zeros in a quadratic polynomial have a direct relation with the coefficients of variables in the polynomial. Then it becomes very easy to calculate the sum and product of zeros even without knowing the values of zeros of a polynomial. Let's learn about it in detail in this lesson.
What are Quadratic Polynomials?
Polynomials are expressions formed by variable(s). 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} ≠ 0.
A quadratic polynomial is an expression with the highest degree 2. It's a degree two polynomial as the highest power of variables is 2.
Zeros of Quadratic 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 linear polynomial: p(x) = x+2. The zero of the polynomial is 2 because when x = 2, p(x)=0. Consider a quadratic polynomial: p(x)= x^{2}−9. The zeros of the polynomial are ±3. When x = ±3, p(x)=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.
A polynomial of degree n will have n number of zeros or roots. Thus, a quadratic polynomial can have at most two zeros, whereas a cubic polynomial can have at most 3 zeros.
Quadratic Polynomial Formula
Quadratic polynomial formula helps us to solve the value of a polynomial with degree 2 easily and quickly. A quadratic polynomial is of the form p(x): ax^{2}+bx+c, where a≠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: \(\begin{align}x =\frac{b \pm \sqrt{b^24ac}}{2a}\end{align}\), in which,
 a is the coefficient of x^{2} and is also called the quadratic coefficient
 b is the coefficient of x and is also called the linear coefficient and
 c is the constant term
Sum of Roots of a Quadratic Polynomial
As you might know that there are two roots of a quadratic polynomial. If α and β are the zeros of the quadratic polynomial f(x): ax^{2}+bx+c, the sum of the roots of the polynomial is: α+β= −b/a. In other words, it refers to (coefficient of x)/ (coefficient of x^{2}).
Product of Roots of a Quadratic Polynomial
If α and β are the zeros of the quadratic polynomial f(x): ax^{2}+bx+c, the product of the roots of the polynomial is: αβ= c/a.
In other words, it refers to (constant term)/ (coefficient of x^{2}).
Let's summarize the sum and product of the roots formula of a quadratic equation through the image below:
Relation between the Coefficient and the Sum and Product of the Zeros
In this section, we will explore the relationship between the coefficient and the sum and product of the zeros of any 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 this relation between the coefficient and the zeros, let’s consider some examples.
Let, a(x): 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: a(x)= (x−1)(x−2), where Zeroes are x=1 and x=2. 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(x): x^{2}−7x+6. The zeroes can be determined by factorizing: b(x)= (x−1)(x−6) ⇒ Zeroes are, x=1, x=6. 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(x): (x−a)(x−b). The zeroes are a and b. Let us denote the sum and product of the zeroes by S and P respectively. S= a+b and P= ab.
Next, we note that the polynomial can be written (by multiplication) as follows: p(x)= x^{2}−ax−bx+ab ⇒ p(x)= x^{2}−(a+b)x+ab ⇒ p(x)=x^{2}−Sx+P
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(x): (x−a)(x−b). 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(x): lx^{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(x): l(x−a)(x−b). 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(x): x^{2}+ex+f, then the sum and product of the zeroes are simply: S= −e/1= −e and P= f/1= f.
Important Notes:
 The quadratic formula to find the roots of a quadratic equation p(x):ax^{2}+bx+c is \(\begin{align}{b \pm \sqrt{b^24ac} \over 2a}\end{align}\).
 The sum of the zeroes of a quadratic polynomial equals the negative of the coefficient of x by the coefficient of x^{2}.
 The product of the zeroes equals the constant term by the coefficient of x^{2}.
 A polynomial having the value 0 is called a zero polynomial.
Solved Examples on Sum and Product of Quadratic Polynomials

Example 1: Consider the following quadratic polynomial p(x): 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, p(x)=x^{2}−13x+42 = x^{2}−6x−7x+42 = x(x−6)−7(x−6) = (x−6)(x−7). The zeroes are 6 and 7. So, the sum of zeros is 6+7 = 13 and the product of zeros are 6×7= 42. Therefore, the sum of the zeros is 13 and their product is 42.

Example 2: The sum and product of the zeroes of a quadratic polynomial p are 9 and 20 respectively. Also, p(6) = 4. Determine the polynomial p(x).
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(x): k(x^{2}−9x+20). Now, p(6)= 4 ⇒ k(6^{2}−9(6)+20)= 4 ⇒ k(36−54+20)= 4 ⇒ 2k= 4 ⇒ k= 2. Note how the extra information enabled us to determine the value of k. Therefore, the polynomial is p(x): 2(x^{2}−9x+20). This implies, p(x): 2x^{2}−18x+40.

Example 3: Consider the following quadratic polynomial p(x): x^{2}+x+1. What is the sum and product of the zeroes of this polynomial?
Solution:
We have: S =  coefficient of x/coefficient of x^{2} =  1/1 =  1 , P = constant/coefficient of x^{2} = 1/1 = 1. Simple enough, but here's 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.
Practice Questions on Sum and Product of Quadratic Polynomials
FAQs on Sum and Product of Quadratic Polynomials
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 −b/a. That is coefficient of x^{2}/coefficient of x^{3}.
How to Find the Zeros of a Polynomial?
Zeros of any polynomial can be found by following the steps given below:
 Step 1: Use the Rational Zero Theorem to list all possible rational zeros of the polynomial. Rational zero of the polynomial = constant term/ leading term.
 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
What are the Zeros of a Polynomial?
The zeroes 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(x)= x^{2}−16. The zeros of the polynomial are ±4. When x= ±4, p(x)=0.
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.
What is the Sum of the Zeros of a Quadratic Polynomial?
The Sum of zeros of a quadratic polynomial given as ax^{2}+bx+c can be found by taking the negative value of the ratio of the coefficient of x by the coefficient of x^{2}. It is given as b/a.
How to Form Quadratic Polynomials with Sum and Product of its Zeros?
Quadratic polynomials can be written as p(x): x^{2} Sx + P, where S is the sum of zeros and P is the product of zeros.
How do you Find the Sum and Product of the Roots of a Quadratic Equation?
For a quadratic equation, ax^{2}+bx+c= 0, the sum of the roots is b/a and the product of the roots is c/a.
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