$f(x+h)=0$ represents the transformed equation of the equation $f(x)=x^4+2 x^3-19 x^2-8 x+60=0$. If this…
$f(x+h)=0$ represents the transformed equation of the equation $f(x)=x^4+2 x^3-19 x^2-8 x+60=0$. If this transformation removes the term containing $x^3$ from $f(x)=0$, then $h=$
$-\frac{1}{2}$
$1$
$2$
$-1$
Solution
Given, $f(x)=x^4+2 x^3-19 x^2-8 x+60=0$
Now, $f(x+h)=0$
$\begin{aligned} & \Rightarrow(x+h)^4+2(x+h)^3-19(x+h)^2-8(x+h)+60=0 \\ & \Rightarrow x^4+4 x^3 h+6 x^2 h^2+4 x h^3+h^4+2 x^3+2 h^3\end{aligned}$
$\begin{aligned} & +6 x h(x+h)-19(x+h)^2-8(x+h)+60\end{aligned}=0$
It does not contain $x^3$ So, $4 h+2=0 \Rightarrow h=\frac{-2}{4}=\frac{-1}{2}$