How do you factor completely \[{{x}^{2}}-12x+35\]?

Hint:As the quadratic polynomial is given, first equate this polynomial with zero and make it an equation. Then compare it with the standard form of a quadratic equation and then calculate the discriminant of the quadratic equation if it is equal to or greater than zero then the equation has real roots means it can be further factorized otherwise it has imaginary roots and not factorized.

Complete solution:
First of all, equate this polynomial with zero
\[\Rightarrow {{x}^{2}}-12x+35=0\]
Comparing this with the standard form of quadratic equation and that is
\[a{{x}^{2}}+bx+c=0\]
On comparing we get
\[a=1\] , \[b=-12\] , \[c=35\]
Now calculating the discriminant\[(D)\]
\[D={{b}^{2}}-4ac\]
From given equation
\[D={{(-12)}^{2}}-4\times 1\times 35\]
On solving,
\[D=4\]
Since, \[D>0\] that means the roots or zeros of this equation are real
\[\Rightarrow \] The equation is further factorized
Now completing the perfect square method
Since \[{{x}^{2}}-2ax+b={{(x-a)}^{2}}+b-{{a}^{2}}\]
\[\Rightarrow {{x}^{2}}-12x+35=0\]
Now adding \[1\] to both sides
\[\Rightarrow {{x}^{2}}-12x+35+1=0+1\]
\[\Rightarrow {{(x-6)}^{2}}=1\]
\[\Rightarrow {{(x-6)}^{2}}-1=0\]
Now using the identity \[{{x}^{2}}-{{y}^{2}}=(x-y)(x+y)\]
\[\Rightarrow (x-6-1)(x-6+1)=0\]
\[\Rightarrow (x-7)(x-5)=0\]
Since we have equate the given polynomial with zero and the above equation is similar to that one that means our polynomial is \[(x-7)(x-5)\]

Hence the factors of \[{{x}^{2}}-12x+35\] are \[(x-7)\] and \[(x-5)\].

Note: In this type of questions in which we need to factorize or need to find the roots first equate this with zero and then calculate the discriminant which says whether the quadratic polynomial is further factorizable or not. We can also use a hit and trial method in which we find two numbers whose sum is \[b\] and the product is \[c\] provided that the coefficient of highest degree is unity.