Page:Elementary algebra (1896).djvu/94

76 Rh 2. If the third term of the trinomial is negative, then the second terms of its factors have opposite signs.

95. The beginner will frequently find that it is not easy to select the proper factors at the first trial. Practice alone will enable him to detect at a glance whether any pair he has chosen will combine so as to give the correct coefficients of the expression to be resolved.

Ex. Resolve into factors 7x^2 - 19x -6.

Write (7x 3) (x 2) for a first trial, noticing that 3 and 2 must have opposite signs. These factors give 7x^ and - 6 for the first and third terms. But since 7 x 2-3 x 1=11, the combination fails to give the correct coefficient of the middle term.

Next try (7x 2) (x 3).

Since 7 8 - 2 1=19, these factors will be correct if we insert the signs so that the negative shall predominate.

Thus 7x^2-19x-6=(7x+ 2)(x- 8). [Verify by mental multiplication.]

96. In actual work it will not be necessary to put down all these steps at length. The student will soon find that the different cases may be rapidly reviewed, and the unsuitable combinations rejected at once.

Ex. 1. Resolve into factors 14x^2+29x-15..... . (1), 14x^2-29x-15. . . (2)

In each case we may write (7x 3) (2x 5) as a first trial, noticing that 3 and 5 must have opposite signs. And since 7 x 5-3 x 2 = 29, we have only now to insert the proper signs in each factor.

In (1) the positive sign must predominate. In (2) the negative sign must predominate.

Therefore 14x^2 + 29x -15=(7x-3)(2x+ 5). 14x^2 - 29x -15=(74+4+3)(2x-5).

Ex. 2. Resolve into factors 5x^2+17x+6 ..... . (1), 5x^2-17x+6 .(2)

In (1) we notice that the factors which give 6 are both positive. In (2) we notice that the factors which give 6 are both negative.