Question

How do you factor \[{{x}^{2}}+3x+4x+12\] by grouping?

Answer 1

Hint: Take x common from the first two terms and 4 common from the last two terms. Now, take (x + 3) common and finally write \[{{x}^{2}}+3x+4x+12\] as a product of two terms given as (x – a) (x – b). Here, ‘a’ and ‘b’ are called zeroes of the polynomial.

Complete step by step solution:Here, we have been asked to factorize the quadratic polynomial \[{{x}^{2}}+3x+4x+12\] by grouping.
Now, as we can see that we already have four terms in the given quadratic expression, this is because the middle term 7x is already split into two terms, 3x and 4x, and here we just have to group them together such that we may get two factors.
As we can see that we can take x common from the first two terms and 4 common from the last two terms, so grouping the first two terms and the last two terms, we get,
\[\Rightarrow {{x}^{2}}+3x+4x+12=x\left( x+3 \right)+4\left( x+3 \right)\]
Now, taking the term (x + 3) common in the R.H.S, we get,
\[\Rightarrow {{x}^{2}}+3x+4x+12=\left( x+3 \right)\left( x+4 \right)\]
On comparing it with the expression (x – a) (x – b), we have a = -3 and b = -4, therefore -3 and -4 are the zeroes of the polynomial.
Hence, \[\left( x+3 \right)\left( x+4 \right)\] is the factored form of the given expression.

Note: One may note that we can also write the given expression as \[{{x}^{2}}+4x+3x+12\] by rearranging the terms 3x and 4x. Now, you can take x common from the first two terms and 3 common from the last two terms. In this way also we will get the same factors and zeros. Note that here the middle term is already split and that is why we were not required to use the middle term split method.