Wikipedia:Reference desk/Archives/Mathematics/2006 October 8

Factoring Trinomials
Hi all,

I'm wondering how I can figure out the possible values of b for factorable trinomials in the form ax2 + bx + c such as:

2x2 + bx + 4

Thanks, Alex Ng 02:08, 8 October 2006 (UTC)


 * I assume that you want b to be an integer, and the factoring to succeed in the rationals. From the factorization in C, we see that it is required that the discriminant b2 – 4ac be a perfect square, say d2. Now b2 – 4ac = d2 iff b2 – d2 = 4ac. We see that b and d must have the same parity, so p = (b + d) / 2 and q = (b – d) / 2 are integers. Now b = p + q and d = p – q, so we have b2 – d2 = (p + q)2 – (p – q)2 = 4pq. The requirement for the discriminant being a perfect square is therefore pq = ac. Each factorization of ac as p × q gives a value p + q for b that results in a factorable quadratic form. For the example, ac = 8, so we have 1 × 8, –1 × –8, 2 × 4, and –2 × –4. This means b can be one of 9, –9, 6, and –6. Since ac always has the factorization a × c, we see that b = a + c is always possible. --Lambiam Talk  03:01, 8 October 2006 (UTC)


 * A simpler way of achieving the same: Write a in the form a1a2 and c in the form c1c2, any which way you like. Put b = a1c2 + a2c1. Then ax2 + bx + c = (a1x + c1)(a2x + c2). --Lambiam Talk  05:58, 8 October 2006 (UTC)


 * Oh! I get it now! Thanks Lambiam for the very complete explanation (and the extra add-on for less-mathematically inclined people like me!)
 * So in summary: to find b in 2x2 + bx + 4: a x c = r.
 * Factor r. (2 x 4 = 8; 8 = 1 x 8, -1 x -8 ... etc), then add the factors (1 + 8 = 9, -1 + (-8) = -9, etc)
 * Thanks again! Alex Ng 19:59, 8 October 2006 (UTC)

If all else fails, just use the quadratic formula to find roots. -- Ķĩřβȳ ♥  Ťįɱé  Ø  09:11, 8 October 2006 (UTC)


 * That quadratic formula seems very confusing, I'll try it out though, thanks Kirbytime! Alex Ng 19:59, 8 October 2006 (UTC)

b is bigger than or equal to 6 * 21/2