Sure, let's break down the process of factoring the trinomial [tex]\( 12v^2 + 11v - 15 \)[/tex].
First, identify the coefficients in the trinomial [tex]\( 12v^2 + 11v - 15 \)[/tex]:
- The coefficient of [tex]\( v^2 \)[/tex] term (a) is 12.
- The coefficient of [tex]\( v \)[/tex] term (b) is 11.
- The constant term (c) is -15.
We want to factor this into the form [tex]\( (pv + q)(rv + s) \)[/tex]. To do this, we need to find numbers p, q, r, and s such that:
1. [tex]\( pr = 12 \)[/tex]
2. [tex]\( qs = -15 \)[/tex]
3. [tex]\( ps + qr = 11 \)[/tex]
From the calculations, we obtain the solution as follows:
1. The factors of 12 are 1, 2, 3, 4, and 6.
2. The factors of -15 are 1, -1, 3, -3, 5, and -5.
By trial and error (which might involve several steps if solving manually):
- We noticed that the correct pairs are [tex]\( p = 3 \)[/tex], [tex]\( q = 5 \)[/tex], [tex]\( r = 4 \)[/tex], and [tex]\( s = -3 \)[/tex].
Putting these together, the factored form of [tex]\( 12v^2 + 11v - 15 \)[/tex] is:
[tex]\[ (3v + 5)(4v - 3). \][/tex]
Therefore, the correct numbers to fill in the blanks are:
[tex]\[ (3v + 5)(4v - 3). \][/tex]
