Question

How do you factor by grouping four-term polynomials and trinomials?

Answer 1

Hint: First we have to discuss the polynomial and trinomial concept by using the examples. Then we need to check if there are any terms in common, if it is we have to group the terms by taking out the commons. If it is not we can split a term into two in order to form a common and proceed. Finally we get the required answer.

Complete step-by-step solution:
First of all, polynomial is an equation or an expression that consists of variables, its coefficient (constants) and exponents that are combined using mathematical operations i.e., addition, subtraction, multiplication and division.
The four term polynomial is nothing but an expression of 4 terms which are to be factored.
Let’s take an example, \[{\text{xy + 8x + 4y}} + 32 = 0\]
So here there are \[\;4\] terms consisting of two variables.
Now we have to group this expression into two. We can group this expression in any way,
\[ \Rightarrow \left( {{\text{xy + 8x}}} \right) + \left( {{\text{4y}} + 32} \right)\]
The Only requirement is that the grouped terms should have something in common.
Now we can write it as, \[\left( {{\text{xy + 8x}}} \right) + \left( {{\text{4y}} + 32} \right) = 0\]
Clearly in the first group, ${\text{x}}$ is common and in second group $4$ is common,
Taking ${\text{x}}$ as common in first term and $4$ as common in second term, we get,
$ \Rightarrow {\text{x}}({\text{y}} + 8) + {\text{4(y + 8) = 0}}$
So now $({\text{y}} + 8)$ is common, taking $({\text{y}} + 8)$ out,
We get, $({\text{y}} + 8)({\text{x}} + 4) = 0$
Now we have to discuss the trinomial,
Trinomial is an expression of three terms containing variables, exponential and constants combined by the mathematical operations (addition, subtraction, multiplication and division).
Let’s take an example
\[{{\text{x}}^2} + 3{\text{x}} - 18 = 0\] is of the form \[{\text{a}}{{\text{x}}^2} + b{\text{x + c}} = 0\]
So we can compare like this we get,
${\text{a}}{{\text{x}}^{\text{2}}}{\text{ = }}{{\text{x}}^{\text{2}}}{\text{,a = 1}}$
${\text{bx = 3x}}$ and
${\text{c}} = 18$
First, we need to multiply ${\text{a}}$ and ${\text{c}}$ $\left( {1 \times 18 = 18} \right)$ and its factors
Factors of $18$ are $(1,18)(2,9)(3,6)$ and check that any of the set has a difference of sum of $b$
\[(3,6)\] is the one that can be used.
Now we can group the equation as,
\[ \Rightarrow {{\text{x}}^2} + 3{\text{x}} - 18 = 0\]
\[ \Rightarrow {{\text{x}}^2} + 6{\text{x - 3x - 18}} = 0\]
Now taking ${\text{x and - 3}}$ as common we get,
\[ \Rightarrow {\text{x}}\left( {{\text{x}} + 6} \right){\text{ - 3}}\left( {{\text{x + 6}}} \right) = 0\]
Taking \[\left( {{\text{x + 6}}} \right)\] as common
\[ \Rightarrow \left( {{\text{x - 3}}} \right)\left( {{\text{x + 6}}} \right) = 0\]
\[ \Rightarrow \left( {{\text{x + 6}}} \right) = 0{\text{ (x - 3) = 0}}\]
\[ \Rightarrow {\text{x = - 6 , x = 3}}\] are the factors of \[{{\text{x}}^2} + 3{\text{x}} - 18 = 0\].
Hence we get the required answer.

Note: In such a type of questions where students need to find the factors or the roots of the given polynomial or trinomial or any other equations, students may go wrong or get confused when using minus signs for splitting the terms.
When there is a minus sign in the given equation, you need to be very careful throughout solving the sum. While taking minus sign out, the terms will change to opposite signs. For example,
 \[ \Rightarrow {{\text{x}}^2} - 2{\text{x - 3x + 6}} = 0\]
Now taking $ - 3$ out, $ + 6$ becomes $ - 6$
\[ \Rightarrow {\text{x}}\left( {{\text{x - 2}}} \right){\text{ - 3}}\left( {{\text{x - 2}}} \right) = 0\]
Taking \[\left( {{\text{x - 2}}} \right)\] as common we get
\[ \Rightarrow \left( {{\text{x - 3}}} \right)\left( {{\text{x - 2}}} \right) = 0\]
Thus, we get the answer.