Is \[{{x}^{2}}-10x+25\] a perfect square trinomial and how do you factor it?

Hint: Consider the coefficient of \[{{x}^{2}}\], coefficient of x and the constant terms as a, b and c respectively. Now, calculate the discriminant (D) given as: - \[D={{b}^{2}}-4ac\], of the given equation. If the discriminant value is 0 then we can say that the given quadratic polynomial is a perfect square trinomial. To factor the equation write it in the form \[{{a}^{2}}-2ab+{{b}^{2}}\] and use the conversion \[{{\left( a-b \right)}^{2}}\] to get the answer.

Complete answer:
Here, we have been provided with the quadratic polynomial: \[{{x}^{2}}-10x+25\] and we are asked to check if it is a perfect square trinomial or not. In the next step we have to factorize it.
Now, the given quadratic polynomial will only be a perfect square trinomial if it can be written in the form \[{{\left( x-m \right)}^{2}}\]. As we can see that both the roots of the equation will be the same if such a condition arises, so the discriminant value must be zero. Considering the coefficient of \[{{x}^{2}}\], coefficient of x and the constant term as a, b and c respectively, we have,
For \[{{x}^{2}}-10x+25\],
\[\Rightarrow \] a = 1
\[\Rightarrow \] b = -10
\[\Rightarrow \] c = 25
Applying the formula of discriminant (D), we have,
\[\begin{align}
  & \Rightarrow D={{b}^{2}}-4ac \\
 & \Rightarrow D={{\left( -10 \right)}^{2}}-4\times 1\times 25 \\
 & \Rightarrow D=100-100 \\
 & \Rightarrow D=0 \\
\end{align}\]
Therefore, the value of discriminant turns out to be 0, so we can say that \[{{x}^{2}}-10x+25\] is a perfect square trinomial.
Now, let us factor the given quadratic polynomial. We can write the given expression as: -
\[\because {{x}^{2}}-10x+25={{x}^{2}}-2\times x\times 5+{{5}^{2}}\]
Clearly, the R.H.S. of the above expression is of the form \[{{a}^{2}}-2ab+{{b}^{2}}\] whose factored form is \[{{\left( a-b \right)}^{2}}\], so we can write,
\[\Rightarrow {{x}^{2}}-10x+25={{\left( x-5 \right)}^{2}}\]
Hence, \[{{\left( x-5 \right)}^{2}}\] is the factored form of the given quadratic expression.

Note: One may note that if the value of D would have been greater than 0 then we would have used the middle term split method or completing the square method to factorize the quadratic polynomial. If the value of D would have been negative, i.e., less than 0, then we cannot factorize the quadratic polynomial. If the value of D would have been negative, i.e., less than 0, then we cannot factorize the polynomial because in that case the roots would not have been real. Remember the formulas: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right),{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}},{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], because they are used everywhere in the topic ‘algebra’.