Zeroes of a Cubic Polynomial

We have seen that cubic polynomials are of the form

\(p\left( x \right):a{x^3} + b{x^2} + cx + d,\,\,a \ne {\rm{0}}\)

Any such polynomial will have, in general, three zeroes. For example, \(p\left( x \right):{x^3} - 6{x^2} + 11x - 6\) has the following three zeroes (verify that these are indeed the zeroes of the polynomial): \(x = 1,\,\,2,\,\,3\).

The three zeroes of a cubic polynomial might all be equal. For example, consider \(p\left( x \right):{\left( {x - 1} \right)^3}\). This has the three zeroes \(x = 1,\,\,1,\,\,1,\) which happen to be identical.

Another case which is possible is that two of the zeroes are equal, and the third is different. For example, consider \(p\left( x \right):{\left( {x - 1} \right)^2}\left( {x - 2} \right)\). This has the three zeroes: \(x = 1,\,\,1,\,\,2\).

Will a cubic polynomial always have three real zeroes? The answer is no. Just as a quadratic polynomial does not always have real zeroes, a cubic polynomial may also not have all its zeroes as real. But there is a crucial difference. A cubic polynomial will always have at least one real zero. Thus, the following cases are possible for the zeroes of a cubic polynomial:

1. All three zeroes might be real and distinct.

2. All three zeroes might be real, and two of them might be equal.

3. All three zeroes might be real and equal.

4. One zero might be real and the other two non-real (complex).

The reasons behind these properties of zeroes will become clear later. For now, you should try to memorize these facts so that things will be easier for you when you reach the stage of studying them in more detail.

Consider the following cubic polynomial, written as the product of three linear factors:

\[p\left( x \right):  \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 4} \right)\]

The zeroes of this polynomial are:

\[x = 1,\;x = 2,\;x = 4\]

The sum and product of the zeroes are:

\[\begin{align}&S = 1 + 2 + 4 = 7\\&P = 1 \times 2 \times 4 = 8\end{align}\]

Now, let us multiply the three factors in the first expression, and write the polynomial in standard form. Then, we will explore what relation the sum and product of the zeroes has with the coefficients of the polynomial:

\[\begin{align}&p\left( x \right) = \underbrace {\left( {x - 1} \right)\left( {x - 2} \right)}_{}\left( {x - 4} \right)\\& = \left( {{x^2} - 3x + 2} \right)\left( {x - 4} \right)\\& = {x^3} - 4{x^2} - 3{x^2}\; + 12x + 2x - 8\\& = {x^3} - 7{x^2} + 14x - 8\end{align}\]

Observe that the coefficient of \({x^2}\) is –7, which is the negative of the sum of the zeroes. The constant term is –8, which is the negative of the product of the zeroes. Now, let us evaluate the sum t of the product of zeroes taken two at a time:

\[\begin{align}&T = 1 \times 2 + 2 \times 4 + 1 \times 4\\&= 2 + 8 + 4\\&= 14\end{align}\]

This is the same as the coefficient of x in the polynomial’s expression. Let us explore these connections more formally. Consider the following cubic polynomial:

\[p\left( x \right):  a{x^2} + bx + cx + d\;\;\;\;...(1)\]

Suppose that this cubic polynomial has three zeroes, say α, β and γ. Then, we can write this polynomial as:

\[p\left( x \right) = a\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\]

Can you see how this can be done? The multiplier of a is required because in the original expression of the polynomial, the coefficient of \({x^3}\) is a. Now, let us expand this product above:

\[\begin{align}&p\left( x \right) = a\underbrace {\left( {x - \alpha } \right)\left( {x - \beta } \right)}_{}\left( {x - \gamma } \right)\\&= a\left( {{x^2} - \left( {\alpha  + \beta } \right)x + \alpha \beta } \right)\left( {x - \gamma } \right)\\&= a\left( \begin{array}{l}{x^3} - \left( {\alpha  + \beta  + \gamma } \right){x^2}\\ + \left( {\alpha \beta  + \beta \gamma  + \alpha \gamma } \right)x - \alpha \beta \gamma \end{array} \right)\\&= a\left( {{x^3} - S{x^2} + Tx - P} \right)\;...\;(2)\end{align}\]

s is the sum of the zeroes, t is the sum of the product of zeroes taken two at a time, and p is the product of the zeroes:

\[\begin{array}{l}S = \alpha  + \beta  + \gamma \\T = \alpha \beta  + \beta \gamma  + \alpha \gamma \\P = \alpha \beta \gamma \end{array}\]

Comparing the expressions marked (1) and (2), we have:

\[\begin{align}&a{x^3} + b{x^2} + cx + d = a\left( {{x^3} - S{x^2} + Tx - P} \right)\\&\Rightarrow \;\;\;{x^3} + \frac{b}{a}{x^2} + \frac{c}{a}x + \frac{d}{a} = {x^3} - S{x^2} + Tx - P\\&\Rightarrow \;\;\;\frac{b}{a} = - S,\;\frac{c}{a} = T,\;\frac{d}{a} = - P\\&\Rightarrow \;\;\;\left\{ \begin{gathered}S = - \frac{b}{a} = - \frac{{{\rm{coeff}}\;{\rm{of}}\;{x^2}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^3}}}\\T = \frac{c}{a} = \frac{{{\rm{coeff}}\;{\rm{of}}\;x}}{{{\rm{coeff}}\;{\rm{of}}\;{x^3}}}\\P = - \frac{d}{a} = - \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^3}}}\end{gathered} \right.\end{align}\]

Thus, we have obtained the expressions for the sum of zeroes, sum of product of zeroes taken two at a time, and product of zeroes, for any arbitrary cubic polynomial. As an example, suppose that the zeroes of the following polynomial are p, q and r:

\[f\left( x \right): 2{x^3} - 12{x^2} + 22x - 12\]

Without even calculating the zeroes explicitly, we can say that:

\[\begin{array}{l}p + q + r =  - \frac{{\left( { - 12} \right)}}{2} = 6\\pq + qr + pr = \frac{{22}}{2} = 11\\pqr =  - \frac{{\left( { - 12} \right)}}{2} = 6\end{array}\]

Example 1: Consider the following polynomial:

\[p\left( x \right): 3{x^3} - 11{x^2} + 7x - 15\]

What is the product of the zeroes of this polynomial?

Solution: We have:

\[P =  - \frac{{{\rm{constant}}}}{{{\rm{coeff}}\;{\rm{of}}\;{x^3}}} =  - \frac{{\left( { - 15} \right)}}{3} = 5\]

Example 2: Determine a polynomial about which the following information is provided:

1. The sum of its zeroes is 12.

2. The sum of the product of its zeroes taken two at a time is 47.

3. The product of its zeroes is 60.

Solution: Given the sum of zeroes (s), sum of product of zeroes taken two at a time (t), and the product of the zeroes (p), we can write a cubic polynomial as:

\[p\left( x \right):  k\left( {{x^3} - S{x^2} + Tx - P} \right)\]

k can be any real number. Its value will have no effect on the zeroes. In this particular case, the answer will be:

\[p\left( x \right):  k\left( {{x^3} - 12{x^2} + 47x - 60} \right)\]

where k can be any real number. However, if an additional constraint is given – for example, if the value of the polynomial is given for a certain x value – then the value of k will also become uniquely determined, as in the following example.

Example 3: Determine the polynomial about which the following information is provided:

1. The sum of its zeroes is 1.

2. The sum of the product of its zeroes taken two at a time is \(- 10\).

3. The product of its zeroes is 8.

4. \(p\left( 0 \right) =  - 24\)

Solution: We can write the polynomial as:

\[\begin{align}&p\left( x \right) = k\left( {{x^3} - \left( 1 \right){x^2} + \left( { - 10} \right)x - \left( 8 \right)} \right)\\&= k\left( {{x^3} - {x^2} - 10x - 8} \right)\end{align}\]

Now, we have:

\[\begin{array}{l}p\left( 0 \right) =  - 24\\ \Rightarrow \;\;\;k\left( { - 8} \right) =  - 24\\ \Rightarrow \;\;\;k = 3\end{array}\]

Thus, the required polynomial is

\[\begin{align}&p\left( x \right) = 3\left( {{x^3} - {x^2} - 10x - 8} \right)\\&= 3{x^3} - 3{x^2} - 30x - 24\end{align}\]

Example 4: Consider the following polynomial:

\[p\left( x \right):  {x^3} - 5{x^2} + 3x - 4\]

What is the sum of the squares of the zeroes of this polynomial?

Solution: Let the zeroes of this polynomial be α, β and γ. We have:

\[\begin{array}{l}\alpha  + \beta  + \gamma  =  - \frac{{\left( { - 5} \right)}}{1} = 5\\\alpha \beta  + \beta \gamma  + \alpha \gamma  = \frac{3}{1} = 3\\\alpha \beta \gamma  =  - \frac{{\left( { - 4} \right)}}{1} = 4\end{array}\]

Now, we make use of the following identity:

\[\begin{array}{l}{\left( {\alpha  + \beta  + \gamma } \right)^2} = \left\{ \begin{array}{l}\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right) + \\2\left( {\alpha \beta  + \beta \gamma  + \alpha \gamma } \right)\end{array} \right.\\ \Rightarrow \;\;\;\;\,\;\;\;  {\left( 5 \right)^2} = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( 3 \right)\\ \Rightarrow \;\;\;\;\,\;\;\;  25 = {\alpha ^2} + {\beta ^2} + {\gamma ^2} + 6\\ \Rightarrow \;\;\;\;\,\;\;\;  {\alpha ^2} + {\beta ^2} + {\gamma ^2} = 19\end{array}\]

Example 5: Consider the following polynomial:

\[p\left( x \right):  2{x^3} - 3{x^2} + 4x - 5\]

What is the sum of the reciprocals of the zeroes of this polynomial?

Solution: Let the zeroes of this polynomial be α, β and γ. We have:

\[\begin{array}{l}\alpha  + \beta  + \gamma  =  - \frac{{\left( { - 3} \right)}}{2} = \frac{3}{2}\\\alpha \beta  + \beta \gamma  + \alpha \gamma  = \frac{4}{2} = 2\\\alpha \beta \gamma  = \;\;\; - \frac{{\left( { - 5} \right)}}{2}\; = \frac{5}{2}\end{array}\]

Now, we have:

\[\begin{align}&\frac{1}{\alpha } + \frac{1}{\beta } + \frac{1}{\gamma } = \frac{{\beta \gamma  + \alpha \gamma  + \alpha \beta }}{{\alpha \beta \gamma }}\\& = \frac{2}{{5/2}}\\&= \frac{4}{5}\end{align}\]

Example 6: Consider the following cubic polynomial:

\(p\left(x\right):\;x^3\;-\;6x^2+\;11x-6\)

Which of the following are zeroes of this polynomial?

(A) \( - 1\)                               (B) \(1\)

(C) \(2\)                                  (D) \(3\)

(E)  \(4\)

Solution: The correct options are (B), (C) and (D). Verify the same by substitution.