Let R denote the set of all real numbers. Let | vedclass

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(C) Given $h(x) = f(x+1) - g(x+2)$.Expanding the terms involving $x^3$ in $h(x)$:$f(x+1) = a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4$The coeff

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$-4$

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(C) Given $h(x) = f(x+1) - g(x+2)$.Expanding the terms involving $x^3$ in $h(x)$:$f(x+1) = a_1 + 10(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + (x+1)^4$The coefficient of $x^3$ in $f(x+1)$ is $a_3 + 4(1) = a_3 + 4$.$g(x+2) = b_1 + 3(x+2) + b_2(x+2)^2 + b_3(x+2)^3 + (x+2)^4$The coefficient of $x^3$ in $g(x+2)$ is $b_3 + 4(2) = b_3 + 8$.Thus,the coefficient of $x^3$ in $h(x)$ is $(a_3 + 4) - (b_3 + 8) = a_3 - b_3 - 4$.Since $f(x) - g(x) 
eq 0$ for all $x \in R$,the expression $f(x) - g(x) = (a_3 - b_3)x^3 + (a_2 - b_2)x^2 + 7x + (a_1 - b_1)$ must not have any real roots.For a cubic polynomial to have no real roots,the coefficient of $x^3$ must be $0$ (otherwise,it would be a cubic equation which always has at least one real root).Therefore,$a_3 - b_3 = 0$.Substituting this into the coefficient of $x^3$ in $h(x)$,we get $0 - 4 = -4$.

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