Question

Factorise $6{x^2} + 17x + 5$ by splitting the middle term and by using the Factor Theorem.

Answer 1

Hint: Here, for the splitting the middle term method, the middle term of the given quadratic function is written in such a way that by taking terms common, we will get the two factors. For Factor Theorem, we will find those values of x which will make the value of the given function equal to zero.

Complete step-by-step answer:

Let the given function in variable x is $f(x) = 6{x^2} + 17x + 5{\text{ }} \to {\text{(1)}}$
By splitting the middle term method, we can write the given function as
$ \Rightarrow f(x) = 6{x^2} + 15x + 2x + 5$
By taking 3x common from the first two terms and 1 common from the last two terms given in the RHS of the above equation, we get
$ \Rightarrow f(x) = 3x\left( {2x + 5} \right) + 1\left( {2x + 5} \right)$
By taking (2x+5) common from the two terms given in the RHS of the above equation, we get
$ \Rightarrow f(x) = \left( {3x + 1} \right)\left( {2x + 5} \right)$
According to equation (1), we can write
So, $6{x^2} + 17x + 5 = \left( {3x + 1} \right)\left( {2x + 5} \right)$
Therefore, the quadratic function in variable x i.e., $6{x^2} + 17x + 5$ have two functions which are (3x+1) and (2x+5).
By Factor Theorem
In order to factorise the given function, those factors are to be identified which will exactly divide the given function leaving zero remainder or those values of x for which the value of the function $6{x^2} + 17x + 5$ comes equal to zero i.e., $f(x) = 0$
Put $x = - \dfrac{1}{3}$ in equation (1), we get
$
   \Rightarrow f\left( { - \dfrac{1}{3}} \right) = 6{\left( { - \dfrac{1}{3}} \right)^2} + 17\left( { - \dfrac{1}{3}} \right) + 5 \\
   \Rightarrow f\left( { - \dfrac{1}{3}} \right) = 6\left( {\dfrac{1}{9}} \right) - \dfrac{{17}}{3} + 5 \\
   \Rightarrow f\left( { - \dfrac{1}{3}} \right) = \dfrac{2}{3} - \dfrac{{17}}{3} + 5 \\
   \Rightarrow f\left( { - \dfrac{1}{3}} \right) = \dfrac{{2 - 17 + 15}}{3} = \dfrac{0}{3} \\
   \Rightarrow f\left( { - \dfrac{1}{3}} \right) = 0 \\
 $
So, $x = - \dfrac{1}{3}$ is the root of the equation $f(x) = 0$.
Hence, $\left( {x + \dfrac{1}{3}} \right) = \left( {\dfrac{{3x + 1}}{3}} \right)$ is one of the factor of the given function i.e., $6{x^2} + 17x + 5$.
Now put $x = - \dfrac{5}{2}$ in equation (1), we get
$
   \Rightarrow f\left( { - \dfrac{5}{2}} \right) = 6{\left( { - \dfrac{5}{2}} \right)^2} + 17\left( { - \dfrac{5}{2}} \right) + 5 \\
   \Rightarrow f\left( { - \dfrac{5}{2}} \right) = 6\left( {\dfrac{{25}}{4}} \right) - \dfrac{{85}}{2} + 5 \\
   \Rightarrow f\left( { - \dfrac{5}{2}} \right) = \dfrac{{75}}{2} - \dfrac{{85}}{2} + 5 \\
   \Rightarrow f\left( { - \dfrac{5}{2}} \right) = \dfrac{{75 - 85 + 10}}{2} = \dfrac{0}{2} \\
   \Rightarrow f\left( { - \dfrac{5}{2}} \right) = 0 \\
 $
So, $x = - \dfrac{5}{2}$ is the root of the equation $f(x) = 0$.
Hence, $\left( {x + \dfrac{5}{2}} \right) = \left( {\dfrac{{2x + 5}}{2}} \right)$ is the other factor of the given function i.e., $6{x^2} + 17x + 5$.
So, the two factors of the given quadratic function $6{x^2} + 17x + 5$ are $\left( {\dfrac{{3x + 1}}{3}} \right)$ and $\left( {\dfrac{{2x + 5}}{2}} \right)$.
We can also say that the two factors of the given quadratic function $6{x^2} + 17x + 5$ are $\left( {3x + 1} \right)$ and $\left( {2x + 5} \right)$.

Note: In this particular problem, we can clearly see that splitting the middle term is a better method as compared to the method of Factor Theorem because in Factor Theorem, we have to take trial cases in order to actually reach to the final values of x will which make the value of the given function equal to zero.