f(x+h)=0 represents the transformed equation of the equation f(x)=x4+2x3−19x2−8x+60=0. If this transformation removes the term containing x3 from f(x)=0, then h=
To find the value of h that removes the x3 term from the polynomial f(x)=x4+2x3−19x2−8x+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 x3 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 x3 Terms: When expanding, consider only the terms involving x3 : From (x+h)4 : ‌4C1⋅x3⋅h=4hx3 From 2(x+h)3 : 2×1×x3=2x3 Collect x3 Terms: Adding these terms together gives: 4hx3+2x3=(4h+2)x3 Set the Coefficient to Zero: To eliminate the x3 term from the polynomial, the coefficient must be zero: 4h+2=0 Solve for h : 4h=−2‌‌⟹‌‌h=‌
−2
4
=‌
−1
2
Thus, the value of h that removes the x3 term is h=−‌