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

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

asked Sep 12, 2014

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.

answered Sep 12, 2014

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.

answered Sep 12, 2014

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.

answered Sep 12, 2014
edited Sep 12, 2014 by david

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.

answered Sep 12, 2014