Question

Find the zeroes of the polynomial \[{x^3} - 5{x^2} - 2x + 24\], it is given that the product of its two zeroes is \[12\].

Answer 1

Hint: Here we will be using the formula for calculating the sum and product of the zeroes of the cubic polynomial which states that for any cubic polynomial
\[a{x^3} + b{x^2} + cx + d = 0\] if the zeroes are \[\alpha \], \[\beta \] and \[\gamma \] then the formula for calculating the sum and product will be as below:
\[{\text{Sum of zeroes}} = \dfrac{{ - {\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}\]
\[{\text{Product of zeroes}} = \dfrac{{ - {\text{constant}}}}{{{\text{coefficient of }}{x^3}}}\]

Complete step-by-step solution:
Step 1: We have the polynomial \[f\left( x \right) = {x^3} - 5{x^2} - 2x + 24\] and let
\[\alpha \], \[\beta \] and \[\gamma \] are the three zeroes of it then as per the given information we can write as below:
 \[\alpha \beta = 12\] ………… (1)
Also, by using the formula of calculating the sum and product of zeroes of a cubic polynomial we can write the below equations:
\[\alpha + \beta + \gamma = \dfrac{{ - {\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}\]
By substituting the value of the coefficient of
\[{x^2}\] and \[{x^3}\] in the above equation, we get:
\[ \Rightarrow \alpha + \beta + \gamma = \dfrac{{ - \left( { - 5} \right)}}{1}\]
By opening the brackets of the RHS side of the above equation we get:
\[ \Rightarrow \alpha + \beta + \gamma = 5\] ……………………… (2)
Similarly, we can write the product formula as below:
\[ \Rightarrow \alpha \beta \gamma = \dfrac{{ - 24}}{1}\]
We can write the above equation as below:
\[ \Rightarrow \alpha \beta \gamma = - 24\] …………………….. (3)
Step 2: By substituting the value from equation (1) in equation (3), we get:
\[ \Rightarrow 12\gamma = - 24\]
Bringing \[12\] into the RHS side of the equation and dividing it with \[ - 24\], we get:
\[ \Rightarrow \gamma = - 2\] ……………………….. (4)
By substituting the value of \[\gamma = - 2\] in the equation (2), we get:
\[ \Rightarrow \alpha + \beta + \left( { - 2} \right) = 5\]
Bringing \[ - 2\] into the RHS side of the above equation, we get:
\[ \Rightarrow \alpha + \beta = 7\] ………………………… (5)
Now, squaring on both the sides of the equation (5), we get:
\[ \Rightarrow {\left( {\alpha + \beta } \right)^2} = {\left( 7 \right)^2}\]
Step 3: By using the formula of the quadratic polynomial
\[{\left( {a + b} \right)^2} = {\left( {a - b} \right)^2} + 4ab\] in the above equation we get:
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta = 49\]
By substituting the value \[\alpha \beta = 12\] from equation (1) in the above equation, we get:
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} + 4\left( {12} \right) = 49\]
By doing multiplication of \[4\left( {12} \right)\] and bringing it in the RHS side of the equation we get:
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} = 49 - 48\]
By doing subtraction on the RHS side of the above equation, we get:
\[ \Rightarrow {\left( {\alpha - \beta } \right)^2} = 1\]
Taking root on both sides, we get:
\[ \Rightarrow \alpha - \beta = 1\] …………………………… (6)
Step 4: By solving equation (5) and (6) and adding them, we get:

\[
  \alpha + \beta = 7 \\
  \alpha - \beta = 1 \\
  \overline {2\alpha + 0 = 8} \\
 \]
From the above equation, by taking
\[2\] into the RHS side of the equation and dividing it with
\[8\], we get:
\[ \Rightarrow \alpha = 4\] …………………………. (7)
By substituting the value of \[\alpha = 4\]into the equation (6), we get:
\[ \Rightarrow 4 - \beta = 1\]
By solving the above equation, we get:
\[ \Rightarrow \beta = 3\]

The zeroes of the cubic polynomial are \[\alpha = 4\],\[\beta = 3\] and \[\gamma = - 2\]

Note: Students need to remember that zeroes of any polynomial are those values of the variable for which the polynomial as a whole has zero value.
For example, for the polynomial \[p\left( x \right) = x + 2\], the zero will equal to \[ - 2\], because at this value the polynomial will become zero.
\[ \Rightarrow p\left( x \right) = - 2 + 2\]
\[p\left( x \right) = 0\]