Question

How do you factor the trinomial \[{x^2} + 2x - 4\]?

Answer 1

Hint: In the given question, we have been asked to factorize the given quadratic equation. To do that, we can easily solve it by using the formula of calculating the value of the variable by using the concept of determinants. We use the formula for solving the value of determinants. Then we just put in the value of determinant into the formula for finding the variable, and that gives us the answer. But it is a point to note that if the value of the determinant is less than 0, i.e., negative, then there is no possible solution for the given equation.

Complete step by step answer:
We are going to use the formula of calculating the value of the variable by using Determinant:
Determinant, \[D = {b^2} - 4ac\]
and, \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
The given quadratic equation is \[{x^2} + 2x - 4\].
First, we calculate the value of the determinant.
In this question, \[a = 1,{\rm{ }}b = 2,{\rm{ }}c = - 4\].
The formula for finding determinant is:
\[D = {b^2} - 4ac\]
Putting in the values, we get,
\[D = {\left( 2 \right)^2} - 4 \times 1 \times \left( { - 4} \right) = 4 + 16 = 20 = {\left( {2\sqrt 5 } \right)^2}\]
Now, we put in the value of the determinant into calculating the value of \[x\],
\[x = \dfrac{{ - 2 \pm \left( {2\sqrt 5 } \right)}}{2} = - 1 \pm 5\]

Hence, \[{x^2} + 2x - 4 = \left( {x + 1 - \sqrt 5 } \right)\left( {x + 1 + \sqrt 5 } \right)\]

Note: In the given question, we had to factorize the given polynomial. The biggest mistake that some students make is to start factoring the polynomial by splitting the middle without checking the determinant and end up wasting a lot of time. If the calculated value of determinant is negative, there is no possible solution. So, before solving by splitting the middle term, we need to check the value of determinant and then solve.