How do you factor a perfect square trinomial \[4{{a}^{2}}-10a-25\] ?

Hint: From the question given perfect square trinomial is \[4{{a}^{2}}-10a-25\]. We have to find the factors for the given perfect square trinomial. Perfect square number means it’s a given number from a number system that can be expressed as the square of a number from the same number system. Here the given trinomial is not a perfect square. So, we have to find the factors by using quadratic formula i.e., $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step by step solution:
From the question given, the trinomial is \[4{{a}^{2}}-10a-25\] . It is not a perfect square trinomial.
Perfect square trinomials are the result of squaring binomials, i.e.,
\[\begin{align}
  & \Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & \Rightarrow {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align}\]
The last number in the given question trinomial is negative i.e., \[-25\]
but in the perfect square trinomial the last number can not be negative because it is a squared number.
By this we can say that the given trinomial is not a perfect square trinomial.
Therefore, we have to find the factor of the given trinomial by using the quadratic formula.
the roots of the quadratic equation$a{{x}^{2}}+bx+c=0$ given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ then if the two solutions are $p,q$ then the factors will be $\left( x-p \right)\left( x-q \right)$ .
here, in the given trinomial
\[\begin{align}
  & \Rightarrow a=4 \\
 & \Rightarrow b=-10 \\
 & \Rightarrow c=-25 \\
\end{align}\]
By substituting the above values in the formula, we will get,
 \[\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\Rightarrow \dfrac{-\left( -10 \right)\pm \sqrt{100+4\left( 100 \right)}}{2\times 4}\]
\[\Rightarrow \dfrac{10\pm 10\sqrt{5}}{8}\]
\[\Rightarrow \dfrac{5\pm 5\sqrt{5}}{4}\]
Therefore, the factors are \[\left( a-\dfrac{5+5\sqrt{5}}{4} \right)\] and \[\left( a-\dfrac{5-5\sqrt{5}}{4} \right)\] for the trinomial equation \[4{{a}^{2}}-10a-25\].

Note:
Students should know the basic definition of perfect square. If in the question the trinomial is \[4{{a}^{2}}-10a+25\] then it is a perfect square trinomial because the last term is also positive and it can be written as
\[\Rightarrow {{\left( 2a-5 \right)}^{2}}=4{{a}^{2}}-10a+25\]
Therefore, the factor of this perfect square trinomial is \[\left( 2a-5 \right)\]