Question

How do you factor $12{{y}^{2}}+7y+1$ ?

Answer 1

Hint: To factor the given expression in the above problem. First of all, we should know the degree of this expression. Degree of any polynomial is the highest power of the variable. After seeing the degree, you will find that the given expression is a quadratic expression and to factorize the given quadratic expression we are going to multiply the coefficient of ${{y}^{2}}$ and the constant and then write the factors for this result of multiplication. And then arrange the factors in such a way so that addition or subtraction will give the coefficient of x.

Complete step by step solution:
The expression given above which we have to solve is as follows:
$12{{y}^{2}}+7y+1$
As you can see that the highest power of this equation is 2 so the degree of the above equation is 2 and hence, the above expression is a quadratic.
Now, we are going to factorize this quadratic expression by finding the factors of 12.
$\begin{align}
  & 12=12\times 1 \\
 & 12=2\times 6 \\
 & 12=3\times 4 \\
\end{align}$
If you look at the last factor then you will find that on adding this factor (3 + 4) you will get the coefficient of y which is 7 so substituting $3+4$ in place of 7 in the above quadratic expression we get:
$\begin{align}
  & 12{{y}^{2}}+\left( 3+4 \right)y+1 \\
 & \Rightarrow 12{{y}^{2}}+3y+4y+1=0 \\
\end{align}$
Taking $3y$ as common from first two terms in the above expression and 1 as common from the last two terms and we get,
$3y\left( 4y+1 \right)+1\left( 4y+1 \right)$
Now, taking $\left( 4y+1 \right)$ as common from the above we get,
$\left( 4y+1 \right)\left( 3y+1 \right)$
Hence, we have factorized the given expression to $\left( 4y+1 \right)\left( 3y+1 \right)$.

Note: You can check the factors solved in the above solution, by equating each factor to 0 and the put then you will get the values of y and then substitute these values of y in the original expression and see if these values are making the expression 0 or not.
The factors which we have solved above are as follows:
$\left( 4y+1 \right)\left( 3y+1 \right)$
Equating each of the brackets to 0 we get,
$\begin{align}
  & 4y+1=0\Rightarrow y=-\dfrac{1}{4}; \\
 & 3y+1=0\Rightarrow y=-\dfrac{1}{3} \\
\end{align}$
Let us check the first value of y by substituting in the original quadratic expression we get,
$\begin{align}
  & 12{{y}^{2}}+7y+1 \\
 & =12{{\left( -\dfrac{1}{4} \right)}^{2}}+7\left( -\dfrac{1}{4} \right)+1 \\
 & =12\left( \dfrac{1}{16} \right)-\dfrac{7}{4}+1 \\
 & =\dfrac{3}{4}+1-\dfrac{7}{4} \\
\end{align}$
Taking 4 as L.C.M in the denominator of the above expression we get,
$\begin{align}
  & =\dfrac{3+4-7}{4} \\
 & =\dfrac{7-7}{4}=0 \\
\end{align}$
This means that the factor $\left( 4y+1 \right)$ is correct. Similarly, you can check the other factor too.