Functions

To find the value of $h$ that removes the $x^3$ term from the polynomial $f(x)=x^4+2x^3-19x^2-8x+60=0$ when transformed to $f(x+h)=0$, follow these steps:Polynomial Transformation: The transformation affects each term in the original polynomial. For the $x^3$ term in the polynomial, apply the transformation:$f(x+h)=(x+h)^4+2(x+h)^3-19(x+h)^2-8(x+h)+60$Focus on the $x^3$ Terms: When expanding, consider only the terms involving $x^3$ :From $(x+h)^4$ :${}^{4}C_{1} \cdot x^{3} \cdot h = 4 h x^{3}$From $2(x+h)^3$ :$2 \times 1 \times x^{3} = 2 x^{3}$Collect $x^3$ Terms: Adding these terms together gives:$4 h x^{3} + 2 x^{3} = (4 h + 2) x^{3}$Set the Coefficient to Zero: To eliminate the $x^3$ term from the polynomial, the coefficient must be zero:$4 h + 2 = 0$Solve for $h$ :$4 h = -2 \qquad \Rightarrow \qquad h = \frac{-2}{4} = \frac{-1}{2}$Thus, the value of $h$ that removes the $x^3$ term is $h = -\frac{1}{2}$.