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(D) Let $f(x) = ax^2 + bx + c$.Given $f(0) + f(1) = 0$,we have $c + (a + b + c) = 0$,which implies $a + b + 2c = 0$ $(i)$.Given $f(-2) = 0$,we have $4

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$f\left(\frac{3}{5}\right)=0$

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(D) Let $f(x) = ax^2 + bx + c$.Given $f(0) + f(1) = 0$,we have $c + (a + b + c) = 0$,which implies $a + b + 2c = 0$ $(i)$.Given $f(-2) = 0$,we have $4a - 2b + c = 0$ (ii).From $(i)$,$b = -a - 2c$. Substituting into (ii):$4a - 2(-a - 2c) + c = 0$ $\Rightarrow 4a + 2a + 4c + c = 0$ $\Rightarrow 6a + 5c = 0$.Let $a = 5k$,then $c = -6k$.Substituting into $(i)$: $5k + b + 2(-6k) = 0$ $\Rightarrow 5k + b - 12k = 0$ $\Rightarrow b = 7k$.Thus,$f(x) = k(5x^2 + 7x - 6) = k(5x^2 + 10x - 3x - 6) = k(5x(x + 2) - 3(x + 2)) = k(5x - 3)(x + 2)$.The roots are $x = -2$ and $x = \frac{3}{5}$.Therefore,$f\left(\frac{3}{5}\right) = 0$.

Let $f(x)$ be a quadratic expression such that $f(0)+f(1)=0$. If $f(-2)=0$,then

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