If x=a is a root of multiplicity two of a pol | vedclass

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(D) Given that $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$. This implies that $f(x)$ can be written as $f(x)=(x-a)^2 g(x)$,w

Vedclass · Accepted Answer

$f(a)=f^{\prime}(a)=0 ; f^{\prime \prime}(a) 
eq 0$

Vedclass · Answer

(D) Given that $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$. This implies that $f(x)$ can be written as $f(x)=(x-a)^2 g(x)$,where $g(a) 
eq 0$. First,we find the first derivative $f^{\prime}(x)$: $f^{\prime}(x) = 2(x-a)g(x) + (x-a)^2 g^{\prime}(x)$. Evaluating at $x=a$,we get $f^{\prime}(a) = 2(a-a)g(a) + (a-a)^2 g^{\prime}(a) = 0$. Next,we find the second derivative $f^{\prime \prime}(x)$: $f^{\prime \prime}(x) = 2g(x) + 2(x-a)g^{\prime}(x) + 2(x-a)g^{\prime}(x) + (x-a)^2 g^{\prime \prime}(x) = 2g(x) + 4(x-a)g^{\prime}(x) + (x-a)^2 g^{\prime \prime}(x)$. Evaluating at $x=a$,we get $f^{\prime \prime}(a) = 2g(a) + 4(a-a)g^{\prime}(a) + (a-a)^2 g^{\prime \prime}(a) = 2g(a)$. Since $g(a) 
eq 0$,it follows that $f^{\prime \prime}(a) 
eq 0$. Thus,$f(a)=0$,$f^{\prime}(a)=0$,and $f^{\prime \prime}(a) 
eq 0$.

If $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$,then

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