If α and β are the zeros of the quadratic polynomial

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-5 x+4$

Therefore $\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-(5)}{1}$

= 5

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{4}{1}$ 

= 4

We have, $\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta$

$\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta=\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta$

$\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta=\frac{\alpha+\beta}{\alpha \beta}-2 \alpha \beta$

By substituting $\alpha+\beta=5$ and $\alpha \beta=4$ we get,

$\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta=\frac{5}{4}-2(4)$

$\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta=\frac{5}{4}-\frac{8 \times 4}{1 \times 4}$

Taking least common factor we get ,

$\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta=\frac{5-32}{4}$

$\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta=\frac{-27}{4}$

Hence, the value of $\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta$ is $\frac{-27}{4}$.