Question

Factor each expression.12) 7k(k-3)+4(k-3)13) 3y(2y+3)-5(2y+3)

Answer 1

`7k(k-3)+4(k-3)`

To factor this expression we need to find the common factor and take it out

In the two terms above the bracket (k - 3) is a common factor, because it is repeated in the two terms, then let us take it

`7k(k-3)+4(k-3)=(k-3)\lbrack7k+4\rbrack`

There is no like term in the bracket [7k + 4], then the answer is

(k - 3)[7k + 4]

13.

`3y(2y+3)-5(2y+3)`

The common factor of the two terms is the bracket (2y + 3), then we will take it

`3y(2y+3)-5(2y+3)=(2y+3)\lbrack3y-5\rbrack`

No like terms in the bracket [3y-5], then

The answer is (2y+3)[3y-5]