The factors of $9{\mathrm{a}}^2-{\mathrm{b}}^2+4\mathrm{b}-4$ are:

Given, $9a^2-b^2+4b-4$ 

Simplifying the given expression, we get

$\Rightarrow 9{\mathrm{a}}^2-{\mathrm{b}}^2+4\mathrm{b}-4$

$\Rightarrow 9{\mathrm{a}}^2-\left({\mathrm{b}}^2-4\mathrm{b}+4\right)$

$\Rightarrow {\left(3\mathrm{a}\right)}^2-\left[{\left(\mathrm{b}\right)}^2-2\times \mathrm{b}\times 2+{\left(2\right)}^2\right]$

Using the identity ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{)}}^{\mathbf{2}}$, we get

Here, x = b and y = 2

$\Rightarrow {\left(3\mathrm{a}\right)}^2-{\left(\mathrm{b}-2\right)}^2$

Using the identity ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{)}$, we get

$\Rightarrow \left(3\mathrm{a}-\left(\mathrm{b}-2\right)\right)\left(3\mathrm{a}+\left(\mathrm{b}-2\right)\right)$

$\Rightarrow \left(3\mathrm{a}-\mathrm{b}+2\right)\left(3\mathrm{a}+\mathrm{b}-2\right)$

Hence, factors of $9{\mathrm{a}}^2-{\mathrm{b}}^2+4\mathrm{b}-4 \mathrm{are}\mathbf{ }\left(3a-b+2\right)\left(3a+b-2\right)$.