Question

How do you factor \[{{x}^{2}}+18x+77\]?

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 same 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 \[{{x}^{2}}+18x+77\], the coefficient of \[{{x}^{2}}\]and constant terms are of the same sign and their product is 77. That is
We have a and c coefficients with the same sign in the polynomial of form \[a{{x}^{2}}+bx+c\].
Hence, we would split 18, which is the coefficient of \[x\], in two parts, whose sum is 18 and product is 77. These are 7 and 11.
So, we write it as
\[\Rightarrow \]\[{{x}^{2}}+18x+77\]
We now split \[18x\] into \[11x\] and \[7x\].
\[\Rightarrow {{x}^{2}}+11x+7x+77\]
We take the terms common in the first 2 terms and last 2 terms.
\[\Rightarrow x(x+11)+7(x+11)\]
Here, we have \[(x+11)\] in common then
\[\Rightarrow (x+7)(x+11)\]
\[\therefore (x+7)(x+11)\] is the required answer.

Note:
If the sign of coefficient of \[{{x}^{2}}\] and constant terms are different, 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.