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

May 1, 2016

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

$3 \left(7 {x}^{2} + 9 x - 10\right)$

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, ( $7 \times 10$) and see if any of them differ by 9. Indeed 14 x 5 fits the bill.

Using the method of grouping gives the following:

$3 \left(7 {x}^{2} + 9 x - 10\right)$ = $3 \left(7 {x}^{2} + 14 x - 5 x - 10\right)$

$\left(7 {x}^{2} + 14 x - 5 x - 10\right) = \left(7 {x}^{2} + 14 x\right) + \left(- 5 x - 10\right)$

$= 7 x \left(x + 2\right) - 5 \left(x + 2\right) = \left(x + 2\right) \left(7 x - 5\right)$

$3 \left(7 {x}^{2} + 9 x - 10\right) = 3 \left(x + 2\right) \left(7 x - 5\right)$