What is the value of b and c in the trinomial? Use the table below to determine the two integers, r and s.

Okay Mfu, let's tackle this one.

3.2.1
The given trinomial is $x^2 + x - 12$.
Comparing this to the standard form $x^2 + bx + c$:
$b = \boxed{1}$
$c = \boxed{-12}$

3.2.2
To determine the two integers $r$ and $s$ such that their product is $c = -12$ and their sum is $b = 1$:

Factors of -12	Product of the two factors	Sum of the two factors	Result
1 and -12	$1 \times (-12) = -12$	$1 + (-12) = -11$	product = -12 but sum $\neq$ 1
-1 and 12	$(-1) \times 12 = -12$	$-1 + 12 = 11$	product = -12 but sum $\neq$ 1
2 and -6	$2 \times (-6) = -12$	$2 + (-6) = -4$	product = -12 but sum $\neq$ 1
-2 and 6	$(-2) \times 6 = -12$	$-2 + 6 = 4$	product = -12 but sum $\neq$ 1
3 and -4	$3 \times (-4) = -12$	$3 + (-4) = -1$	product = -12 but sum $\neq$ 1
-3 and 4	$(-3) \times 4 = -12$	$-3 + 4 = 1$	product = -12 and sum = 1

The two integers are $\boxed{-3 \text{ and } 4}$.

3.2.3
Let's complete the table with the mathematical steps for factorizing $x^2 + x - 12$:

Mathematical steps	Explanation
$x^2 + 4x - 3x - 12$	Rewrite the middle term of the trinomial using the values from the chart above.
$(x^2 + 4x) + (-3x - 12)$	Group pairs of terms.
$x(x + 4)$	Factor out the HCF of the first group.
$-3(x + 4)$	Factor out the HCF of the second group.
$(x + 4)(x - 3)$	Factor out HCF of the two terms
$(x + 4)(x - 3)$	The final factorized answer