Question

If two of the zeros of a Cubic polynomial are zero, then it does not have linear and constant
terms.
(A) true
(B) false

Answer 1

Hint :- To solve this, consider a cubic polynomial and then consider three roots and solve
Let $a{x^2} + bx + c = 0$ be any Cubic equation.
and this equation has three roots.
Let $\alpha ,\beta ,\gamma $ are the three roots of this equation.

Complete step by step by solution:
Let we have a Cubic equation with root $\alpha ,\beta ,\gamma $
$f(x) = a{x^2} + bx + c + d = 0$
then we can write in form
$f(x) = (x - \alpha )(x - \beta )(x - \gamma )$
According to the question,
\[\beta = 0,\gamma = 0\]
Put in $f(x)$
$f(x) = (x - \alpha )(x - 0)(x - 0)$
$ \Rightarrow f(x) = (x - \alpha ){x^2}$
$ \Rightarrow f(x) = {x^3} - \alpha {x^2}$
Now we Can See $f(x) $does not have linear and constant terms.
So, the statement of the question is True.


Note –There exists some relation between the roots of the cubic polynomial
$\alpha + \beta + \gamma = - \dfrac{b}{a}$
$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}$
$\alpha \beta \gamma = - \dfrac{d}{a}$
By putting the value of
$\beta = 0 and \gamma = 0$
We find that,
$c = 0$
$d = 0$
and then we have $a{x^3} + b{x^2} = 0$