Question

How do you factor \[7{{x}^{2}}+5x-2\]?

Answer 1

Hint: For factoring polynomials, “factoring” (or “factoring completely”) is always done using some set of numbers as a possible coefficient. When all the factors are linear, it is clear that the expression is factored completely. If we have an irrational factor then we will definitely have its conjugate form also.

Complete step by step answer:
As per the given question, we have to factor \[7{{x}^{2}}+5x-2\] completely.
Firstly, we multiply the coefficient of \[{{x}^{2}}\] with the constant term -2. That is, we get \[\Rightarrow 7\times -2=-14\].
As the product is negative, then we have to choose two numbers whose sum (or difference) is coefficient of x and product of these two numbers must be equal to -14.
Finding the numbers: As we know that the sum of 7 and -2 gives the coefficient of x (that is, 5) and their product is -14.
\[\begin{align}
  & \Rightarrow 7-2=5 \\
 & \Rightarrow 7\times -2=-14 \\
\end{align}\]
So, we can say that the numbers are 7 and -2.
Now, we substitute these values into the expression. Then, we get
\[\Rightarrow 7{{x}^{2}}+7x-2x-2\] (\[\Leftarrow \] Notice that we replaced the 5 with sum of 7 and -2)
Factoring the first two expressions, we can take \[7x\] as common. Then we can write as
\[\Rightarrow 7{{x}^{2}}+7x\to 7x(x+1)\]
Factoring the next two expressions, we can take -2 as common. Then we can write as
\[\Rightarrow -2x-2\to -2(x+1)\]
From above, we notice that (x+1) is the same in both. Therefore, we can rewrite the given expression as:
\[\Rightarrow 7x(x+1)-2(x+1)\to (7x-2)(x+1)\]
Here, we can see that factors are linear. Thus, this is the completely factored form of the given expression \[7{{x}^{2}}+5x-2\].
We can always check by expanding:
\[\Rightarrow (7x-2)(x+1)=7{{x}^{2}}+7x-2x-2\to 7{{x}^{2}}+5x-2\] which is the original. So, we are correct.

\[\therefore \] On factoring the expression \[7{{x}^{2}}+5x-2\] completely, we get \[(7x-2)(x+1)\].

Note: While solving these types of problems, we generally make a mistake leaving the complex factors like \[({{x}^{2}}+4)\] unfactorized. We have to factor these ones also. And, we need to have enough knowledge on the methods of factoring an expression completely. It is better if we avoid calculation or simplification mistakes.