Question

How do you factor the expression ${{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}$ ?

Answer 1

Hint: We are given a polynomial expression as ${{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}$ . We are asked to find the factor of it, to do so first we see that the equation is in fraction so we change it to the integers. So we multiply by appropriate integers, then we learn about the type of equation. we will further find the factor of it using the middle term split in which we will find b (middle term) of $a{{x}^{2}}+bx+c$ by a pair of number such that those terms product is same as $a\times c$ .

Complete step-by-step answer:
We are given an equation as ${{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}$.
We can see that they have a rational, coefficient.
So, we will simplify it to get an easier equation.
We can see that coefficients are $1,\dfrac{2}{3},\dfrac{1}{9}$ .
So, we will multiply and divide terms by ‘9’.
So, we have ${{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}$.
Now by multiplying and divide it by ‘9’, we get –
$=\dfrac{9}{9}\times \left( {{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9} \right)$ .
By simplifying, we get –
$=\dfrac{1}{9}\left( 9{{x}^{2}}+6x+1 \right)$ …………………………………… (1)
Now we will factor $9{{x}^{2}}+6x+1$ by using the middle term split method.
In this method, we will find pairs of numbers such that the difference is the same as ‘b’.
Now in our equation we have $9{{x}^{2}}+6x+1$ .
$a=9,b=6,c=1$ .
$a\times c=9\times 1=9$ and $b=6$ .
We can see that there are numbers 3 and 3.
Such that $3\times 3=9$ (just like $a\times c$ )
And their sum $3+3=6$ .
So, we will use it to factor our middle term.
Now $9{{x}^{2}}+6x+1$ .
We use $6=3+3$ so,
$\Rightarrow 9{{x}^{2}}+\left( 3+3 \right)x+1$
$\Rightarrow 9{{x}^{2}}+3x+3x+1$ (By opening bracket)
Taking common in first two terms and last two terms, we get –
$=3x\left( 3x+1 \right)+1\left( 3x+1 \right)$ .
As $3x+1$ is the same, so take it out.
$\left( 3x+1 \right)\left( 3x+1 \right)$ .
So, we get –
$9{{x}^{2}}+6x+1=\left( 3x+1 \right)\left( 3x+1 \right)$ .
Using this in equation (1), we get –
${{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{1}{9}\left( 3x+1 \right)\left( 3x+1 \right)$ .
So, factor form of ${{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{1}{9}\left( 3x+1 \right)\left( 3x+1 \right)$ .

Note: While finding the middle term we need to be careful that when sign of ‘a’ and ‘c’ are the same then ‘b’ is obtained by sum of the number while when their signs are different then ‘b’ is obtained by difference of terms . we will also remember that we cannot multiply just by numerator or denominator while factoring because doing so will filter the equation and hence it will change our answer, we need to multiply both numerator and denominator then it won’t affect the equation.