How do you factor the trinomial $9{{t}^{2}}-25$?

Hint: We first try to explain the concept of factorisation and the ways a factorisation of a polynomial can be done. We use the identity theorem of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factor the given polynomial $9{{t}^{2}}-25$. We assume the values of $a=3t;b=5$. The final multiplied linear polynomials are the solution of the problem.

Complete step by step solution:
The main condition of factorisation is to break the given number or function or polynomial into multiple of basic primary numbers or polynomials.
For the process of factorisation, we use the concept of common elements or identities to convert into multiplication form.
For the factorisation of the given quadratic polynomial $9{{t}^{2}}-25$, we apply the factorisation identity of difference of two squares as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We get $9{{t}^{2}}-25={{\left( 3t \right)}^{2}}-{{5}^{2}}$. We put the value of $a=3t;b=5$.
Factorisation of the polynomial gives us
$9{{t}^{2}}-25={{\left( 3t \right)}^{2}}-{{5}^{2}}=\left( 3t+5 \right)\left( 3t-5 \right)$.
These two multiplied linear polynomials can’t be broken any more.
Therefore, the final factorisation of $9{{t}^{2}}-25$ is $\left( 3t+5 \right)\left( 3t-5 \right)$.
Therefore, we get $\left( 3t+5 \right)\left( 3t-5 \right)=0$. Multiplied form of two polynomials gives 0 which gives individual terms to be 0.
Therefore, either $\left( 3t+5 \right)$ is 0 or $\left( 3t-5 \right)$ is 0.
The solutions are $t=\pm \dfrac{5}{3}$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have $9{{t}^{2}}-25=0$. The values of a, b, c is $9,0,-25$ respectively.
We put the values and get $x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 9\times \left( -25 \right)}}{2\times 9}=\dfrac{\pm \sqrt{900}}{18}=\dfrac{\pm 30}{18}=\pm \dfrac{5}{3}$.

Note: We find the value of x for which the function $f\left( t \right)=9{{t}^{2}}-25=0$. We can see $f\left( \dfrac{5}{3} \right)=9{{\left( \dfrac{5}{3} \right)}^{2}}-25=25-25=0$. So, the root of the $f\left( t \right)=9{{t}^{2}}-25$ will be the function $\left( t-\dfrac{5}{3} \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Now, $f\left( t \right)=9{{t}^{2}}-25=\left( 3t+5 \right)\left( 3t-5 \right)$. We can also do the same process for $\left( t+\dfrac{5}{3} \right)$.