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
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.
Your answers are correct. Well done! Each step is correct and you have correctly applied the zero-product property to solve for the variables. Your solutions for each problem are accurate. Keep up the good work!
Your solutions to the factoring problems are correct! Here's a breakdown of how you arrived at each solution:
1. z(z - 1)(z + 3) = 0:
To find the solutions, you applied the zero-product property, which states that if a product of factors is equal to zero, then at least one of the factors must be zero. You found the solutions by setting each factor equal to zero and solving for z: z = 0, z - 1 = 0 (which gives z = 1), and z + 3 = 0 (which gives z = -3).
2. x^2 - x - 10 = 2:
First, you rearranged the equation by subtracting 2 from both sides to get x^2 - x - 12 = 0. Then, you factored the quadratic equation into (x - 4)(x + 3) = 0. Applying the zero-product property, you set each factor equal to zero and solved for x, giving x = 4 and x = -3 as the solutions.
3. 4a^2 - 11a + 6 = 0:
You factored the quadratic equation into (a - 2)(4a - 3) = 0. Using the zero-product property, you set each factor equal to zero and solved for a: a - 2 = 0 (which gives a = 2) and 4a - 3 = 0 (which gives a = 3/4).
4. 9r^2 - 30r + 21 = -4:
After rearranging the equation by subtracting -4 from both sides, you obtained 9r^2 - 30r + 25 = 0. Factoring the quadratic equation, you got (3r - 5)^2 = 0. Applying the zero-product property, you set 3r - 5 equal to zero and solved for r, giving r = 5/3 as the solution.
Well done! Your answers correctly utilize factoring and the zero-product property to find the solutions to these equations.