Question

How do you factor ${{n}^{2}}+4n-12$?

Answer 1

Hint: Now we have a quadratic equation in n. to find the factors of the equation we will first split the middle term such that the product of the term is the multiplication of the first term and the last term of the quadratic equation. Now simplify the equation by taking n common from the first two and -2 common from last two terms. Now we will use distributive property to simplify the equation further. Hence we have the factors of the equation.

Complete step by step solution:
Now we are given with a quadratic equation in n.
To factorize the given equation we will use splitting the middle term method.
Now in this method we will split the middle term such that the product of the terms gives us the multiplication of first term and last term.
Now here the middle term is 4n, the first term is ${{n}^{2}}$ and the last term is – 12.
Now split 4n as 6n – 2n. We know that $\left( 6n \right)\times \left( -2n \right)=-12{{n}^{2}}=-12\times {{n}^{2}}$
Now rewriting the equation we get, ${{n}^{2}}+6n-2n-12$ .
Now in the above equation we will take n common from first two terms and -2 common from last two terms. Hence we get,
$\Rightarrow n\left( n+6 \right)-2\left( n+6 \right)$
Now on simplifying the above equation using distributive property we get,
$\Rightarrow \left( n-2 \right)\left( n+6 \right)$
Hence we get the factors of the given equation as $\left( n-2 \right)$ and $\left( n+6 \right)$ .
Note: Now note if $x-\alpha $ and $x-\beta $ are the factors of the equation then we can see that on substituting the value of x as $\alpha $ or $\beta $ we get the value of quadratic as 0. Hence $\alpha $ and $\beta $ are the roots of quadratic equations. Hence we can also write the factors if we know the roots of the equation.