Question

Factorize $2{n^2} + 5n + 2$.

Answer 1

Hint:Quadratic Formula: $a{x^2} + bx + c = 0$ Here $a,\;b,\;c$ are numerical coefficients.So to solve $x$ we have:$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. So in order to solve the above given question using a quadratic formula we have to find the values of $a,\;b,\;c$ corresponding to the given question. But instead of $x$ we have to solve for $n$. Then by substituting the values in the above equation we can find the values for$n$and thereby solve it.

Complete step by step answer:
Given, \[2{n^2} + 5n + 2...............................\left( i \right)\]. Now we need to compare (i) to the general formula and find the values of unknowns. Then we have to use the equation to find $n$ by substituting all the values needed in it and by that way we can solve the equation $2{n^2} + 5n + 2$.So on comparing (i) to the general formula $a{x^2} + bx + c = 0$, we get:
$a = 2,\;b = 5,\;c = 2.....................\left( {ii} \right)$
Now to solve for $n$we have $n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}......................\left( {iii} \right)$
Now substituting (ii) in (iii) we get:
\[
n = \dfrac{{ - \left( 5 \right) \pm \sqrt {{{\left( 5 \right)}^2} - 4\left( {2 \times 2} \right)} }}{{2 \times 2}} \\
\Rightarrow n = \dfrac{{ - 5 \pm \sqrt {\left( {25} \right) - 4\left( 4 \right)} }}{4} \\
\Rightarrow n = \dfrac{{ - 5 \pm \sqrt {\left( {25} \right) - \left( {16} \right)} }}{4} \\
\Rightarrow n = \dfrac{{ - 5 \pm \sqrt 9 }}{4} \\
\Rightarrow n = \dfrac{{ - 5 \pm 3}}{4} \\ \]
Now there are two possibilities of$n$, which is produced either by addition or by subtraction. It’s found such that:
\[
n = \dfrac{{ - 5 + 3}}{4}\;\;\;\;\;{\text{and}}\;\;\;n = \dfrac{{ - 5 - 3}}{4} \\
\Rightarrow n = \dfrac{{ - 2}}{4}\;\;\;\;\;\;\;\;{\text{and}}\;\;\;n = \dfrac{{ - 8}}{4} \\
\Rightarrow n = - \dfrac{1}{2}\;\;\;\;\;\;\;\;\;\;{\text{and}}\;\;\;n = - 2 \\
\Rightarrow n = - 2,\; - \dfrac{1}{2} \\ \]
Now on factoring we get:
$
n = - 2,\; - \dfrac{1}{2} \\
\therefore n = \left( {n + 2} \right)\left( {n + \dfrac{1}{2}} \right) \\ $
Therefore on factoring $2{n^2} + 5n + 2$ we get $\left( {n + 2} \right)\left( {n + \dfrac{1}{2}} \right)$ as the factors.

Additional Information:
Another technique for factoring is grouping, where we take common terms not from the full polynomial but only from the certain terms which have them. So it is used mainly when there is not a common factor for the whole polynomial but there is a common factor only for certain terms.

Note:While approaching a question one should study it properly and accordingly should choose the method to factorize the polynomial. Similar questions which have no proper linear factors with real coefficients should be approached using the same method as described above. Polynomial factorization is always done over some set of numbers which may be integers, real numbers or complex numbers.