Factorize :

x² + x - 12

= x² - 4x + 3x -12

= x( x - 4 ) - 3 ( x - 4 )

= ( x - 3 ) ( x - 4 ) Answer

Since $(ax + b)(cx + d) = acx^2 + adx + cbx + bd$

$acx^2 = x^2, adx + cbx = x$ and $bd = -12$

$acx^2 = x^2 \Rightarrow ac = 1$

If something factorises it means that $a$ and $c$ are both integers which means that $a = 1$ and $c = 1$

$adx + cbx = x \Rightarrow ad + cb = 1$

Since $a, c = 1$

$ad + cb = 1 \Rightarrow d + b = 1$

Rearrange $d + b = 1$ to get an equation for $b$

$b = 1 - d$

Next substitute that equation into $bd = -12$

$(1 - d)d = -12$

The expand and re-arrange

$d - d^2 = -12 \Rightarrow -d^2 + d + 12 = 0 \Rightarrow d^2 - d - 12 = 0$

Since it's equal to zero we can use the quadratic formula

$d = \frac {1 \pm \sqrt {1 - 4(1)(-12)}}{2}$

$d = \frac {1 \pm 7}{2}$

$d = 4, d = -6$

Put these two values back into the equations, we'll start with $d = -6$

$b = 1 - (-6) \Rightarrow b = 1 + 6 = 7$

$b(-6) = -12 \Rightarrow -6b = -12 \Rightarrow b = 2$

As you can see $d \neq -6$

So $d = 4$

$b = 1 - (4) = -3$

$b(4) = -12 \Rightarrow b = -3$

So putting the values back into the original equation gives us

$(x + 4)(x - 3)$

Just obsever, 1 should be the sum of the roots and -12 should be yheir product. Out of the option only $\color{#69047E}{\boxed{(x + 4)(x - 3)}}$ satisfies this condition.

since x=4 x-3 x, so,
x^2+x-12 =>
x^2+4 x-3 x-12 =>
x (x+4)-3 (x+4),
taking commen (x-4),
(x+4)*(x-3)

$-12=4*-3$ and the result follows.

x^2+x-12=x^2+4x-3x-12=x(x+4)-3(x+4)=(x+4)(x-3)