Question

(i) Find the sum of zeros of the cubic polynomial \[2{x^3} - 13{x^2} + 23x - 12\]
(ii) Find the product of the zeros of cubic polynomial \[2{x^3} + 5{x^2} - 4x + 3\]

Answer 1

Hint: Here the question is related to the cubic polynomial. The sum of zeros of the cubic equation is equal to the negative coefficient of \[{x^2}\]by the coefficient of \[{x^3}\]. The product of zeros of the cubic polynomial is equal to the constant by the coefficient of \[{x^3}\]. Hence, we obtain the solution for the given question

Complete step by step answer:
The equation is a polynomial equation and the given equation is a cubic equation.
Now consider the first equation \[2{x^3} - 13{x^2} + 23x - 12\]
The general cubic equation is given as \[a{x^3} + b{x^2} + cx + d\]
By comparing the equation to the general equation, the value of a is 2, the value of b is -13, the value
c is 23 and the value of d is -12.
The sum of zeros of the cubic equation \[2{x^3} - 13{x^2} + 23x - 12\] is given by \[\dfrac{{ - b}}{a}\] , by substituting the values we get it as \[\dfrac{{ - ( - 13)}}{2}\]
on simplifying we get \[\dfrac{{13}}{2}\]
Hence the sum of zeros of the cubic equation \[2{x^3} - 13{x^2} + 23x - 12\] is \[\dfrac{{13}}{2}\]
Now consider the second equation \[2{x^3} + 5{x^2} - 4x + 3\], By comparing the equation to the general equation, the value of a is 2, the value of b is 5, the value of c is -4 and the value of d is 3
The product of zeros is given by \[\dfrac{d}{a}\], by substituting values we get it as \[\dfrac{3}{2}\]
Hence the product of zeros of the cubic equation \[2{x^3} + 5{x^2} - 4x + 3\] is \[\dfrac{3}{2}\].

Note: The sum of zeros is defined by the formula, the formula will depend on the form of the equation and the same thing will also. We should know about sign conventions which are related to the multiplication. If it simplifies for further we can use the division method and we simplify it