Question 190084
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You can go through the process of finding all of the possible factors of -15 and 10, blah blah blah, but there is a much simpler and quicker method.  Use the idea that if *[tex \LARGE x = a] is a root of the equation formed when you set the polynomial equal to zero, then *[tex \LARGE x - a] must be a factor of the polynomial.  Since this is a quadratic trinomial, you can set it equal to zero and use the quadratic formula to find the roots -- then the factors become obvious.

Step 1:  Set the trinomial equal to zero.

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  10w^2-19w-15=0]

Step 2: Solve with the quadratic formula:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w = \frac{-b \pm sqrt{b^2 - 4ac}}{2a} = \frac{-(-19) \pm sqrt{(-19)^2 - 4(10)(-15)}}{2(10)} = \frac{19 \pm sqrt{961}}{20} = \frac{19 \pm 31}{20} ]

So

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  w = \frac{19 + 31}{20} = \frac{50}{20} =\frac{5}{2} \ \ \Rightarrow\ \ 2w - 5 = 0]

Or

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  w = \frac{19 - 31}{20} = \frac{-12}{20} =\frac{-3}{5} \ \ \Rightarrow\ \ 5w + 3 = 0]

Verifying that

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2w - 5)(5w + 3) = 10w^2-19w-15]

is left as an exercise for the student.

By the way, if you end up with a pair irrational roots, then the trinomial is not factorable over the rational numbers.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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