all correct. Good work.
Lots of students don't understand why we always set the polynomial equal to zero to find solutions. You seem to understand that quite well.
I was asked to use factoring and the zero-product property to solve the following problems. Can you please check my answers? Thank you.
z(z - 1)(z + 3) = 0 is already factored, so one of the factors must be zero for the product to be zero. Taking them in turn:
z = 0
z - 1 = 0
z = 1
z + 3 = 0
z = -3
so the three solutions are z = 0, z = 1 and z = -3
x^2 - x - 10 = 2
Arrange in standard form by combining like terms:
x^2 - x - 12 = 0
Factor:
(x - 4) (x + 3) = 0
Apply the zero product principle:
x - 4 = 0
x = 4
x + 3 = 0
x = -3
So the two solutions are x = 4 and x = -3
4a^2 - 11a + 6 = 0
Factor:
(a - 2) (4a - 3) = 0
Apply the zero product principle:
a - 2 = 0
a = 2
4a - 3 = 0
4a = 3
a = 3/4
So the two solutions are a = 2 and a = 3/4
9r^2 - 30r + 21 = -4
Arrange in standard form by combining like terms:
9r^2 - 30r + 25 = 0
Factor:
(3r - 5)^2 = 0
Apply the zero product principle:
3r - 5 = 0
3r = 5
r = 5/3 and r = 5/3
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