Question

if two zeros of cubic polynomial p x is equals to x cube + b x square + c x + d are 0 , then c=d=0 true or false​

Answer 1

Answer:

duzuzddzusddusudzzdudsudsufzufzsidisf

Answer 2

Answer:

No it's false

Step-by-step explanation:

actually the given question also consist zeros 1 and -1 actually it is not given in question given by you

Given

$p(x) = {x}^{3} + bx ^{2} + cx + d$

Substituting 1 as the root of p(x) we get as follows

$p(1) = 0$

$p( - 1) = 0$

hence substituting these values in starting equation we get as follows

$p(1) = ({1})^{3} + b ({1})^{2} + c(1) + d = 0$

$p(1) = 1 + b + c + d = 0$

let it be equation (1)

now taking -1 as other root

$p( - 1) = ({ - 1})^{3} + b ({ - 1})^{2} + c( - 1) + d = 0$

$p( - 1) = ( - 1) + b - c + d = 0$

=>consider it as equation (2)

=>now by considering equation (1) and (2)

=>We get as follows

$p(1) = 0\\ p( - 1) = 0 \\ p(1) + p( - 1) = 0$

$(1 + b + c + d) = 0 \\ ( - 1 + b - c + d )= 0$

adding them we get

$2b + 2d = 0 \\ b + d = 0$

since b+d=0

=>from equation (1) we get

=>c=-1

=>since we got c=-1

=>c≠d

=>so the given statement is false

=>the true statement is

=>c=d=-1

= as b+d=0

By trail and error method

=>b can be any positive or negative integer

=>d can be any positive or negative integer

=>as c=-1 c must be not equal to o

=>hence c≠d≠0