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

seo-qna
Last updated date: 13th Jun 2024
Total views: 374.7k
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Answer
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Hint: To factor $81{c^2} + 198c + 121$ , we have to look at the coefficients of $c$ , ${c^2}$ and the constant part. At first, we have to find out such two numbers which are the multiplication of the factors of $\;81$ and $\;121$ that the addition of those two numbers is equal to $\;198$ . They are $\;99$ and $\;99$ . Then we get some common parts from the first two and second two terms. From this, we can construct the factors.

Complete step by step answer:
We have given;
$81{c^2} + 198c + 121$
The coefficient of ${c^2}$ is $\;81$ and the coefficient of $c$ is $\;121$ . Now $\;81$ is square of $9$ and $\;121$ is the square of $\;11$ . So, we can construct two numbers from the multiplication of the factors of $\;81$ and $\;121$ whose summation will be equal to $\;198$ . $\;99$ and $\;99$ are those two numbers. So we can rewrite the above term as;
$= 81{c^2} + (99 + 99)c + 121$
We can rewrite this as;
$= 81{c^2} + 99c + 99c + 121$
This is clear that $\;9c$ is common in the first two terms and $\;11$ is common in the next two terms. So, we will take $\;9c$ common from the first two terms and $\;11$ common from the last two terms and get;
$= 9c(9c + 11) + 11(9c + 11)$
We can rewrite this as;
$= (9c + 11)(9c + 11)$
We can rewrite this as;
$= {(9c + 11)^2}$
So, our final answer is ${(9c + 11)^2}$ .

Note: To factorize this kind of quadratic equation in the form $a{x^2} + bx + c = 0$ we have to look at the sign before the constant part. If there is a positive sign, the coefficient $x$ will be the summation of two numbers. If there is a negative sign before the constant part, we have to find such two numbers whose subtraction is equal to the coefficient of $x$ .
Alternative Method:
We have given;
$81{c^2} + 198c + 121$
Now $\;81$ is a square of $9$ and $\;121$ is the square of $\;11$ . We can also rewrite $\;198$ as $2 \times 99$ . We will get;
$= {\left( {9c} \right)^2} + 2 \cdot 9c \cdot 11 + {11^2}$
As we know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ we can rewrite the above as;
$= {\left( {9c + 11} \right)^2}$