f(x+h)=0 represents the transformed equation | vedclass

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(A) Given the equation $f(x) = x^4 + 2x^3 - 19x^2 - 8x + 60 = 0$. To remove the $x^3$ term,we substitute $x$ with $(x+h)$. The term containing $x^3$ i

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$-\frac{1}{2}$

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(A) Given the equation $f(x) = x^4 + 2x^3 - 19x^2 - 8x + 60 = 0$. To remove the $x^3$ term,we substitute $x$ with $(x+h)$. The term containing $x^3$ in the expansion of $(x+h)^4 + 2(x+h)^3 - 19(x+h)^2 - 8(x+h) + 60 = 0$ is obtained from the binomial expansion. $(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$. $2(x+h)^3 = 2(x^3 + 3x^2h + 3xh^2 + h^3) = 2x^3 + 6x^2h + 6xh^2 + 2h^3$. The coefficient of $x^3$ in the transformed equation is $4h + 2$. For the $x^3$ term to be removed,we set the coefficient to zero: $4h + 2 = 0$. $4h = -2$. $h = -\frac{2}{4} = -\frac{1}{2}$.

$f(x+h)=0$ represents the transformed equation of the equation $f(x)=x^4+2x^3-19x^2-8x+60=0$. If this transformation removes the term containing $x^3$ from $f(x)=0$,then $h=$

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