f(x)=8x2-19x-15.
Let f(x)=(ax-b)(cx+d), so we have to find a, b, c, d.
Here's how we do it.
Factors of 8=(a,c)={(1,8), (2,4), (4,2), (8,1)}; factors of 15=(b,d)={(1,15), (3,5), (5,3), (1,15)}.
We have (ax-b)(cx+d)=acx2+x(ad-bc)-bd, where ac=8 and bd=15. We want ad-bc=-19.
The table below shows all possibilities. The red row gives us the answer.
f(x)=(x-3)(8x+5), zeroes are x=-⅝ and 3, because either 8x+5=0, 8x=-5, x=-⅝; or x-3=0, x=3.
You can see that this red row is similar to the second to last row in the table but the + and - signs are reversed. We could have used a table half the size and then sorted out the signs (which would always be different) because the sign of the constant term, -15, is negative.
a |
b |
c |
d |
ad |
bc |
ad-bc |
1 |
1 |
8 |
15 |
15 |
8 |
7 |
1 |
3 |
8 |
5 |
5 |
24 |
-19 |
1 |
5 |
8 |
3 |
3 |
40 |
-37 |
1 |
15 |
8 |
1 |
1 |
120 |
-119 |
2 |
1 |
4 |
15 |
30 |
4 |
26 |
2 |
3 |
4 |
5 |
10 |
12 |
-2 |
2 |
5 |
4 |
3 |
6 |
20 |
-14 |
2 |
15 |
4 |
1 |
2 |
60 |
-58 |
4 |
1 |
2 |
15 |
60 |
2 |
58 |
4 |
3 |
2 |
5 |
20 |
6 |
14 |
4 |
5 |
2 |
3 |
12 |
10 |
2 |
4 |
15 |
2 |
1 |
4 |
30 |
-26 |
8 |
1 |
1 |
15 |
120 |
1 |
119 |
8 |
3 |
1 |
5 |
40 |
3 |
37 |
8 |
5 |
1 |
3 |
24 |
5 |
19 |
8 |
15 |
1 |
1 |
8 |
15 |
-7 |