Use factoring and the zero product property to solve these problems. Please help.?

Question

Show all steps for the following problems: 

1.) z(z-1)(z+3)=0 

2.) x^2-x-10=2 

3.) 4a^2-11a+6=0 

4.) 9r^2-30r+21=-4 

Answer 1

1) The equation z ( z - 1) (z + 3) = 0

Apply the zero product property ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0).

z = 0, ( z - 1)= 0 and (z + 3) = 0

• z = 0
• z - 1 = 0

Add 1 to each side.

z = 1

• z + 3 = 0

Subtract 3 from each side.

z = - 3

Solutions are z = 0 ,  z = 1 and z = - 3.

Answer 2

2) The quadratic equation x² - x - 10 = 2

Subtract 2 from each side.

x²  - x - 12 = 0

Now factorize the above equation.

Multiply first term x² and last term - 12 = - 12x²

The correct pair of the terms - 4x and 3x multiply to - 12x² and add to - x.

Replace the middle term -x with - 4x + 3x.

 x² - 4x + 3x - 12 = 0

Group the terms into two pairs.

 (x² - 4x) + (3x - 12) = 0

Factor out x from the first group  and factor out 3 from the second group.

 x(x - 4) + 3(x - 4) = 0

Factor out common term x - 4.

 (x - 4)(x + 3) = 0

Apply the zero product property ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0).

(x - 4) = 0 and (x + 3) = 0

Solutions are x = 4 and x = -3.

Answer 3

3) The equation 4a² - 11a + 6 = 0

Now factorize the above equation.

Multiply first term 4a² and last term  6 =  24a²

The correct pair of the terms - 8a and - 3a multiply to  24a² and add to - 11a.

Replace the middle term - 11a with - 8a - 3a.

 4a² - 8a - 3a + 6 = 0

Group the terms into two pairs.

 (4a² - 8a) + (- 3a + 6) = 0

Factor out 4a from the first group  and factor out - 3 from the second group.

 4a(a - 2) - 3(a - 2) = 0

Factor out common term a - 2.

 (a - 2)(4a - 3) = 0

Apply the zero product property ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0).

(a - 2) = 0 and (4a - 3) = 0

a = 2 and 4a = 3

Solutions are a = 2 and a = 3/4.

Answer 4

4) The quadratic equation 9r² - 30r + 21 = - 4

Add 4 to each side.

9r² - 30r + 25 = 0

(3r)² - 2 (3r) (5) + (5)² = 0

Apply the formula ( a - b)² = a² + b² - 2ab.

In this case a = 3r , b = 25

( 3r - 5)² = 0

(3r - 5)(3r - 5) = 0

Apply the zero product property ab = 0 then a = 0 or b = 0 (or both a = 0 and b = 0).

3r - 5 = 0 or 3r - 5 = 0

Solution  r = 5/3.