Question

How do you factor \[2{{x}^{2}}-7x-4\]?

Answer 1

Hint: When we have a polynomial of the form \[a{{x}^{2}}+bx+c\], we can factor this quadratic with splitting up the b term into two terms based on the signs of a and c terms. If we have the opposite sign for both a and c terms, then we split the b term into two parts. These parts are formed such that the sum of them is b term itself and their product is the same as that of the product of a and c terms.

Complete step by step answer:
In the given quadratic polynomial \[2{{x}^{2}}-7x-4\], the coefficient of \[{{x}^{2}}\] and constant terms are of the opposite sign and their product is -8. That is we have a and c coefficients with opposite signs in the polynomial of form \[a{{x}^{2}}+bx+c\].
Hence, we would split -7, which is the coefficient of x, in two parts, whose sum is -7 and product is -8. These are -8 and 1.
So, we write it as
\[\Rightarrow 2{{x}^{2}}-7x-4\]
We now split \[-7x\] into \[-8x\] and \[1x\].
\[\Rightarrow 2{{x}^{2}}-8x+1x-4\]
We take the terms common in first 2 terms and last 2 terms
\[\Rightarrow 2x(x-4)+1(x-4)\]
Here, we have \[(x-4)\] in common then
\[\Rightarrow (2x+1)(x-4)\]

\[\therefore (2x+1)(x-4)\] is the required answer.

Note: If the coefficient of \[{{x}^{2}}\] and constant terms are the same, then we factor the polynomial by splitting up the b term into two terms. Here also we have to split b term such that sum of those parts is b term and product is same as that of product of a and c terms. Factoring by grouping will not always work. In such cases we better go with the quadratic formula.