How do you factor completely #21x^2 + 27x - 30#?

1 Answer
May 1, 2016

Answer:

Always look for a common factor in all the terms first.

Explanation:

3 is a factor of 21, 27 and 30. Taking it out gives

#3(7x^2 + 9x -10)#

The quadratic trinomial inside the bracket seems not to factorise further, because there are no factor combinations of 7 and 10 which subtract to give 9. The only possibilities are 4, 9, 33 and 69. However, another approach would be to find the factors of 70, ( #7 xx 10#) and see if any of them differ by 9. Indeed 14 x 5 fits the bill.

Using the method of grouping gives the following:

#3(7x^2 + 9x -10)# = #3(7x^2 + 14x -5x -10)#

# (7x^2 + 14x -5x -10) = (7x^2 + 14x) + (-5x -10)#

#=7x(x+2) -5(x+2) = (x+2)(7x-5)#

The final answer would be
#3(7x^2 + 9x -10) = 3(x+2)(7x-5)#