How do you factor \[{{x}^{2}}+18x+81\]?

Hint: Use the middle term split method to factorize \[{{x}^{2}}+18x+81\]. Split 18x into two terms in such a way that their sum equals 18x and product equals \[81{{x}^{2}}\]. To do this, write the prime factors of 81 and group them in such a way that the conditions get satisfied. Finally, take the common terms together and write \[{{x}^{2}}+18x+81\] as a product of two terms given as (x – m) (x – n). Here, ‘m’ and ‘n’ are called zeroes of the polynomial.

Complete step by step solution:
Here, we have been asked to factorize the quadratic polynomial: \[{{x}^{2}}+18x+81\].
Let us use the middle term split method for the factorization. According to this method we have to split the middle term which is 18x into two terms such that their sum is 18x and product is \[81{{x}^{2}}\]. To do this, first we need to find all the prime factors of 81.
We know that 81 can be written as: - \[81=3\times 3\times 3\times 3\] as the product of its primes. Now, we have to group these factors such that the conditions of the middle term split method are satisfied. So, we have,
(i) $\left( 9x \right)+\left( 9x \right)=18x$
(ii) \[\left( 9x \right)\times \left( 9x \right)=81{{x}^{2}}\]
Hence, both the conditions of the middle term split method are satisfied. So, the quadratic polynomial can be written as: -
\[\begin{align}
  & \Rightarrow {{x}^{2}}+18x+81={{x}^{2}}+9x+9x+81 \\
 & \Rightarrow {{x}^{2}}+18x+81=x\left( x+9 \right)+9\left( x+9 \right) \\
\end{align}\]
Taking (x + 9) common in the R.H.S, we have,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+18x+81=\left( x+9 \right)\left( x+9 \right) \\
 & \Rightarrow {{x}^{2}}+18x+81={{\left( x+9 \right)}^{2}} \\
\end{align}\]
Hence, \[{{\left( x+9 \right)}^{2}}\] is the factored form of the given quadratic polynomial.

Note: Here, as you can see we have both the factors of the quadratic polynomial equal to (x + 9), this is because the given expression is of the form of algebraic expression \[{{a}^{2}}+2ab+{{b}^{2}}\], where a = x and b = 9, whose factored form is given directly using the identity \[{{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}\]. So, this can also be a method to factorize the given quadratic polynomial.