Question

How do you factor $81{c^2} + 198c + 121$?

Answer 1

This problem deals with factoring the given expression in $c$. This can be done either by the method of completing the square or just factoring and solving the quadratic equation. To solve $a{x^2} + bx + c = 0$, expression in $x$, by completing the square: transform the equation so that the constant term, $c$ is alone on the right side. But here we are adding and subtracting some terms in order to factor.

Complete step-by-step answer:
Given the quadratic expression is $81{c^2} + 198c + 121$, consider it as given below:
$\Rightarrow 81{c^2} + 198c + 121$
Now expressing the above expression such that the $c$ term is split into half as shown below:
$\Rightarrow 81{c^2} + 99c + 99c + 121$
Now taking the number $9c$ common from the first two terms, and taking the number 11 common from the second two terms, which is shown below:
$\Rightarrow 81{c^2} + 99c + 99c + 121$
$\Rightarrow 9c\left( {9c + 11} \right) + 11\left( {9c + 11} \right)$
Now taking the term $\left( {9c + 11} \right)$ common in the above expression, as shown below:
$\Rightarrow \left( {9c + 11} \right)\left[ {9c + 11} \right]$
$\Rightarrow \left( {9c + 11} \right)\left( {9c + 11} \right)$
So here we factorized the given quadratic expression into two factors, which is shown below:
$\Rightarrow 81{c^2} + 198c + 121 = \left( {9c + 11} \right)\left( {9c + 11} \right)$
One of the factor of $81{c^2} + 198c + 121$ is $\left( {9c + 11} \right)$ and the other factor is also the same as the first factor which is $\left( {9c + 11} \right)$

Final Answer: The factors of $81{c^2} + 198c + 121$ are $\left( {9c + 11} \right)$ and $\left( {9c + 11} \right)$.

Note:
Please note that this problem can also be solved by another method, which is described here. Instead of first factoring and then solving for $x$, we can directly the value of $x$ from the given equation $64{x^2} - 1 = 0$, this can be done by sending the constant 1 to the right hand side of the equation and then solve for $x$, and then factorize with the obtained solutions.