Let f(x)=ax^{2}+bx+c and g(x)=px^{2}+qx+r suc | vedclass

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(C) Let $h(x) = f(x) - g(x) = (a-p)x^2 + (b-q)x + (c-r)$.Since $f(1) = g(1)$,we have $h(1) = 0$.Since $f(2) = g(2)$,we have $h(2) = 0$.Since $h(x)$ is

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$6$

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(C) Let $h(x) = f(x) - g(x) = (a-p)x^2 + (b-q)x + (c-r)$.Since $f(1) = g(1)$,we have $h(1) = 0$.Since $f(2) = g(2)$,we have $h(2) = 0$.Since $h(x)$ is a quadratic polynomial with roots $1$ and $2$,we can write $h(x) = k(x-1)(x-2)$ for some constant $k$.We are given $f(3) - g(3) = 2$,which means $h(3) = 2$.Substituting $x=3$ into $h(x) = k(x-1)(x-2)$,we get $h(3) = k(3-1)(3-2) = k(2)(1) = 2k$.Since $h(3) = 2$,we have $2k = 2$,which implies $k = 1$.Thus,$h(x) = 1(x-1)(x-2) = (x-1)(x-2)$.We need to find $f(4) - g(4)$,which is $h(4)$.$h(4) = (4-1)(4-2) = (3)(2) = 6$.

Let $f(x)=ax^{2}+bx+c$ and $g(x)=px^{2}+qx+r$ such that $f(1)=g(1)$,$f(2)=g(2)$ and $f(3)-g(3)=2$. Then,$f(4)-g(4)$ is:

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