Factor the following trinomial completely. \[ 16 \mathrm{a}^{2}-40 \mathrm{a}+25 \]

To factor the trinomial \( 16 \mathrm{a}^{2}-40 \mathrm{a}+25 \) completely, we first recognize that it resembles the form of a perfect square trinomial. We can check if it can be expressed as \((ma - n)^2\). The first term, \( 16 \mathrm{a}^{2} \), is \( (4\mathrm{a})^2 \), and the last term, \( 25 \), is \( 5^2 \). Next, we check the middle term: \( -40\mathrm{a} \). In a perfect square trinomial, this term should equal \( 2 \cdot (4\mathrm{a}) \cdot (5) = 40\mathrm{a} \) (noting the negative). Since this is true, we can write the trinomial as: \[ (4\mathrm{a} - 5)^2 \] So, the completely factored form is: \[ (4\mathrm{a} - 5)(4\mathrm{a} - 5) \] or simply: \[ (4\mathrm{a} - 5)^2 \]