Question

How do you factor $h\left( x \right) = {x^3} + 2{x^2} - 5x - 6$ ?

Answer 1

Hint: In this question, we will use the concept of the middle term splitting for the quadratic equation. First, we will split the terms such that we will obtain some common factor which will lead to the linear and quadratic equation, then use the concept of the middle term splitting to obtain the factors.

Complete step by step solution:
In this question, we have given a function $h\left( x \right) = {x^3} + 2{x^2} - 5x - 6$, we need to provide the factor of the given function.
In the given function, first we will split $5x$ as $\left( {6x - x} \right)$ and the expression become,
$ \Rightarrow h\left( x \right) = {x^3} + 2{x^2} - \left( {6x - x} \right) - 6$
Now, as we know that the product of positive and the negative number is always a negative number while the product of two negative number is always a positive number, so apply this concept in the above expression and obtain,
$ \Rightarrow h\left( x \right) = {x^3} + 2{x^2} - 6x + x - 6$
Now, we will split $2{x^2}$ as $\left( {{x^2} + {x^2}} \right)$ and the expression become,
$ \Rightarrow h\left( x \right) = {x^3} + {x^2} + {x^2} - 6x + x - 6$
Now, we will make the group of two terms like,
$ \Rightarrow h\left( x \right) = \left( {{x^3} + {x^2}} \right) + \left( {{x^2} + x} \right) - \left( {6x + 6} \right)$
From the above expression we can see that we can take ${x^2}$ as common from the first term, $x$ from the second term and $6$ from the third term combination, so the expression become,
$ \Rightarrow h\left( x \right) = {x^2}\left( {x + 1} \right) + x\left( {x + 1} \right) - 6\left( {x + 1} \right)$
Now, we will take $\left( {x + 1} \right)$ as the common from each term,
$ \Rightarrow h\left( x \right) = \left( {x + 1} \right)\left( {{x^2} + x - 6} \right)$
Now, we will split $x$ term as $\left( {3x - 2x} \right)$ from the second factor of the expression as,
\[ \Rightarrow h\left( x \right) = \left( {x + 1} \right)\left( {{x^2} + 3x - 2x - 6} \right)\]
Now, we will form the combination of the terms as,
\[ \Rightarrow h\left( x \right) = \left( {x + 1} \right)\left( {\left( {{x^2} + 3x} \right) - \left( {2x + 6} \right)} \right)\]
From the above expression we can see that we can take $x$ as common from the first term and $2$ from the second term combination, so the expression become,
\[ \Rightarrow h\left( x \right) = \left( {x + 1} \right)\left( {x\left( {x + 3} \right) - 2\left( {x + 3} \right)} \right)\]
Now, we will take $\left( {x + 3} \right)$ as the common from each term,
\[\therefore h\left( x \right) = \left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 2} \right)\]

Therefore, the factors of the given expression are \[h\left( x \right) = \left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 2} \right)\].

Note:
As we know that the factors of the quadratic equation can be calculated by splitting the middle term such that the sum of the numbers is equal to the coefficient of the $x$ while the product of the number should be equal to the coefficients of ${x^2}$ and the constant.