Question

How do you factor $2{{n}^{2}}+3n-9$?

Answer 1

In this type of question, we should have a good knowledge in polynomials and quadratic equations. In this question, we will first see all the terms and coefficients of the given equation. After that, we will multiply the quadratic term that is $2{{n}^{2}}$ and the constant term that is -9 to get the sum as a linear term as 3n. After that, we will find the pairs of the multiplied terms. From there, we can find the factors by putting the pairs in the place of linear term.

Complete answer:
In this question, we have to factorize the term or we can say we have to find the factors of $2{{n}^{2}}+3n-9$.
Here, the quadratic term is $2{{n}^{2}}$, linear term is 3n and constant term is -9.
Now, we will multiply $2{{n}^{2}}$ and -9 and write all factors of the multiplication result in pairs.
Factors of $2{{n}^{2}}\times (-9)=-18{{n}^{2}}$ will be
-2n and 9n
2n and -9n
18n and -n
-18n and n
-6n and 3n
6n and -3n
From the above pairs, we will check addition of which is equal to 3n which is linear term.
We want to make 3n. For this, we will take 6n and -3n because addition of them is 3n and multiplication of them is $-18{{n}^{2}}$.
Now, from the given equation $2{{n}^{2}}+3n-9$, we will split the linear term 3n into two terms that is 6n and –3n. We get the equation as
$2{{n}^{2}}+6n-3n-9=\left( 2{{n}^{2}}+6n \right)-\left( 3n+9 \right)$
Look at the first set, there is something common in the two numbers of the first set. It is 2n. If we take out the factor 2n, then we will get the first set as 2n(n+3). And, for the second set, we can assume that 3 is common there. So, we can write the second set as 3(n+3).
So, we can write the above equation as
$2n\left( n+3 \right)-3\left( n+3 \right)$
Because both the sets have common a factor that is (n+3), so the above equation will be equal to
(2n-3)(n+3).
Hence, the factor of $2{{n}^{2}}+3n-9$ is $\left( 2n-3 \right)\left( n+3 \right)$.

Note: We have an alternate method to solve this question. For that, we should have a better knowledge in quadratic equations. By using the formula $n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ in the equation $2{{n}^{2}}+3n-9$ where a=2, b=3, and c=-9, we get
$n=\dfrac{-3\pm \sqrt{{{3}^{2}}-4\times 2\times (-9)}}{2\times 2}$
$\Rightarrow n=\dfrac{-3\pm \sqrt{9-(-72)}}{4}$
$\Rightarrow n=\dfrac{-3\pm \sqrt{81}}{4}=\dfrac{-3\pm 9}{4}$
$\Rightarrow n=\dfrac{-3+9}{4}$ and $n=\dfrac{-3-9}{4}$
$\Rightarrow n=\dfrac{6}{4}$ and $n=\dfrac{-12}{4}$
$\Rightarrow n=\dfrac{3}{2}$ and $n=-3$
$\Rightarrow 2n=3$ and $n+3=0$
Hence, factors are (2n-3) and (n+3).
So, we can use this method too.