Question

Given that one of the zeros of the cubic polynomial \[a{x^3} + b{x^2} + cx + d\] is zero, the product of the other two zeros is
A \[ - \dfrac{c}{a}\]
B \[\dfrac{c}{a}\]
C \[0\]
D \[ - \dfrac{b}{a}\]

Answer 1

Hint: In the above question we are given with a cubic polynomial that is \[a{x^3} + b{x^2} + cx + d\] and also we are provided with another condition that is the one of the zeros of the polynomial (zeros are the roots of the equation) is zero we have to find the product of the other two zeros which can be done by firstly putting the one of the zeros which we are given as zero and equating the resultant with zero to further simplify and later on solving for the product of other two roots of the polynomial .

Complete step-by-step answer:
We are given with a cubic polynomial that is \[a{x^3} + b{x^2} + cx + d\] and zero being one of it roots that means by putting \[x = 0\] we should get answer of the polynomial as zero
So putting \[x = 0\]in above polynomial \[a{x^3} + b{x^2} + cx + d\] we get-
\[a{(0)^3} + b{(0)^2} + c(0) + d = 0\]
\[d = 0\]------------\[1\]
So by putting \[x = 0\] in the polynomial \[a{x^3} + b{x^2} + cx + d\] we found out that \[d = 0\]
Now placing this value of \[d = 0\] in the given polynomial whose products of the roots has been asked that is \[a{x^3} + b{x^2} + cx + d\] so the resulting thing becomes as follows –
\[a{x^3} + b{x^2} + cx\]
Now we know that there are three roots of any polynomial equation of the form\[a{x^3} + b{x^2} + cx + d\] are represented as-
\[\alpha ,\beta ,\gamma \]
And also \[\alpha + \beta + \gamma = \dfrac{{ - b}}{a}\]where \[b\] is the coefficient of the term associated with the \[{x^2}\]
And \[a\] is the coefficient of the term associated with the \[{x^3}\] of the polynomial
Also the relation between sum of the product of the roots of the cubic polynomial taken two at a time is as-
 \[\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}\]
Where \[c\]is the coefficient of the term associated with the\[x\] of the polynomial
And \[a\]is the coefficient of the term associated with the \[{x^3}\] of the polynomial
Also the relation between the product of the roots of the polynomial is \[\alpha \beta \gamma = \dfrac{{ - d}}{a}\] where \[d\] is the coefficient of the term associated with the \[x\] and \[a\]is the coefficient of the term associated with the \[{x^3}\] of the polynomial
Now in the question we are one root of the polynomial \[a{x^3} + b{x^2} + cx + d\] is \[0\]so let the \[\alpha \]be zero and using the relation between sum of the product of the roots of the cubic polynomial taken two at a time is as-
 \[\Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}\]
Now putting the \[\alpha \] as \[0\] in above equation we get-
\[\Rightarrow (0)\beta + \beta \gamma + \gamma (0) = \dfrac{c}{a}\]
\[\Rightarrow \beta \gamma = \dfrac{c}{a}\]
Hence we have found the product of the other two roots of the polynomial \[a{x^3} + b{x^2} + cx + d\]
So \[\beta \gamma = \dfrac{c}{a}\] matches with the option B from the above option that is B \[\dfrac{c}{a}\]
So the correct option is B.

Note: while solving such type of questions one should read the statement of the question properly to derive the required facts and molding them accordingly to get the relations and one should be aware of the relations of the sum and products of the roots of the cubic polynomial which makes the question easy to solve .The above question can also be solved by first taking the help of the one root to get the value of \[d\] and then taking \[x\] common and eliminating that \[x\] to make a cubic polynomial into a equation and then using the relation of sum and products of quadratic equation we can solve this question also that is the sum of the roots of quadratic equation of the form \[a{x^2} + bx + c = 0\] is \[\dfrac{{ - b}}{a}\] where \[b\] is the coefficient of the term associated with the \[{x^{}}\] and the a is the coefficient of the term associated with the \[{x^2}\] and product of the roots of quadratic equation is \[\dfrac{c}{a}\] where \[c\] is the constant term.