The idea is to write the factors as follows: (x + ...) (x + ...) (since all terms are positive in this case), then look for numbers that, when added, give 7, and when multiplied, give 10. Possible factor combinations of 10 are 1x10, and 2x5. The only pair of factors that works (that adds up to 7) is 2 and 5.
(x+8)(x-5)
(x + 8)(x - 5)
(5x - 1)(x + 6)
It is x^2 -4 = (x-2)(x+2) when factored and it is the difference of two squares
That trinomial is unfactorable (the roots are not integers).
x2 + 18x + 45 = (x + 15)(x + 3).
(x+8)(x-5)
(x + 8)(x - 5)
(x + 2)(x - 9)
(5x - 1)(x + 6)
(x + 2)(x - 8)
(x + 2)(x - 7)
-((x + 2)(x - 9))
(x - 3)(x + 6)
(5x + 2)(x + 1)
-70
(X+9)(x-2)