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

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Aicharya · Accepted Answer

Given that, $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$. Let $f(x)=(x-a) g(x)$ On differentiating w.r.t. $x$, we get $\Rightarrow f^{\prime}(x)=2(x-a) g(x)+(x-a)^2 g^{\prime}(x)$ Again, differentiating w.r.t $x$, we get $\begin{aligned} & 	ext { Now, } \begin{aligned} & f^{\prime \prime}(x)=2 g(x)+2(x-a) g^{\prime}(x)+2(x-a) g^{\prime}(x) \ &+(x-a)^2 g^{\prime \prime}(x) \end{aligned} \ & =2 g(x)+4(x-a) g^{\prime}(x)+(x-a)^2 g^{\prime \prime}(x) \end{aligned}$ At $x=a$, $\begin{aligned} & \Rightarrow \quad f^{\prime}(a)=2(a-a) g(a)+(a-a)^2 g^{\prime}(a)=0 \ & \Rightarrow \quad f^{\prime \prime}(a)=2 g(a)+4(a-a) g^{\prime}(a)+(a-a)^2 g^{\prime \prime}(a) \ & \quad=2 g(a) \end{aligned}$ Hence, $f(a)=f^{\prime}(a)=0, f^{\prime \prime}(a) 
eq 0$

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

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