Let f(x) = x^3 + a x^2 + b x + c be a polynom | vedclass

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(B) Given the polynomial $f(x) = x^3 + a x^2 + b x + c$. Let the roots of the polynomial be $s - t$,$s$,and $s + t$,where $s$ and $t$ are integers. Ac

Vedclass · Accepted Answer

$1214$

Vedclass · Answer

(B) Given the polynomial $f(x) = x^3 + a x^2 + b x + c$. Let the roots of the polynomial be $s - t$,$s$,and $s + t$,where $s$ and $t$ are integers. According to Vieta's formulas,the sum of the roots is given by: $(s - t) + s + (s + t) = -a$. Simplifying this,we get $3s = -a$,or $a = -3s$. Since $s$ is an integer,$a$ must be a multiple of $3$. Checking the given options: $A: -642 = 3 	imes (-214)$ (Multiple of $3$) $B: 1214$ (Not a multiple of $3$,as $1+2+1+4 = 8$) $C: 1323 = 3 	imes 441$ (Multiple of $3$) $D: 1626 = 3 	imes 542$ (Multiple of $3$) Therefore,'$a$' cannot take the value $1214$.

Let $f(x) = x^3 + a x^2 + b x + c$ be a polynomial with integer coefficients. If the roots of $f(x)$ are integers and are in Arithmetic Progression,then '$a$' cannot take the value:

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