(3r+2)(r-1)= -(7r-7)

Question

I need to factor this quadratic equation then check by substitution

Answer 1

Problem: (3r+2)(r-1)= -(7r-7)
I need to factor this quadratic equation then check by substitution

(3r + 2)(r - 1) = -(7r - 7)
3r^2 - 3r + 2r - 2 = -7r + 7
3r^2 - r - 2 = -7r + 7
3r^2 - r - 2 + 7r = -7r + 7 + 7r
3r^2 + 6r - 2 = 7
3r^2 + 6r - 2 - 7 = 7 - 7
3r^2 + 6r - 9 = 0
(3r - 3)(r + 3) = 0

To check it, you need to solve for r. Set both factors equal to zero and solve.

3r - 3 = 0
3r = 3
r = 1

r + 3 = 0
r = -3

Substitute each of those values into your original equation to make sure they work.

(3r + 2)(r - 1) = -(7r - 7)
(3(1) + 2)(1 - 1) = -(7(1) - 7)
(3 + 2)(1 - 1) = -(7 - 7)
(5)(0) = -(0)
0 = 0

(3r + 2)(r - 1) = -(7r - 7)
(3(-3) + 2)(-3 - 1) = -(7(-3) - 7)
(-9 + 2)(-3 - 1) = -(-21 - 7)
(-7)(-4) = -(-28)
28 = 28

Both answers work.

Answer: r = 1 and r = -3