Use the sum-product pattern
2
−
−
1
2
x
2
−
x
−
12
x2−x−12
2
+
3
−
4
−
1
2
x
2
+
3
x
−
4
x
−
12
x2+3x−4x−12
2
Common factor from the two pairs
2
+
3
−
4
−
1
2
x
2
+
3
x
−
4
x
−
12
x2+3x−4x−12
(
+
3
)
−
4
(
+
3
)
x
(
x
+
3
)
−
4
(
x
+
3
)
x(x+3)−4(x+3)
3
Rewrite in factored form
(
+
3
)
−
4(+3)x
(x+3)−4(x+3)
x(x+3)−4(x+3)
(−4)(+3)
(x−4)(x+3)
(x−4)(x+3)
I think it's proportional
True. For example, if 2 is a, and b was 0, then ab would automatically be 0. Both a and b can be 0 though, but at least one of them must be zero.
Factorise to (5z-1)(z-4) = 0
then put 5z-1 = 0 so 5z = 1 z = 1/5
z-4=0 z = 4
z = 1/5 and 4
I know u gonna use calculator girl.