Answer:
To solve the equation 5r^2 = 36r + 32 using factoring, we'll first rearrange it to make it equal to zero:
5r^2 - 36r - 32 = 0
Now, let's factor this quadratic equation. We're looking for two numbers that multiply to the product of the leading coefficient (5) and the constant term (-32), which is -160, and add up to the coefficient of the linear term (-36).
Let's find two numbers that satisfy these conditions. We can start by listing pairs of factors of -160:
1. (-1, 160)
2. (-2, 80)
3. (-4, 40)
4. (-5, 32)
5. (-8, 20)
6. (-10, 16)
Out of these pairs, we see that the pair (-8, 20) satisfies the condition because -8 * 20 = -160 and -8 + 20 = 12, which is not equal to -36.
So, we can rewrite the middle term (-36r) as -8r + 20r:
5r^2 - 8r + 20r - 32 = 0
Now, let's factor by grouping:
(5r^2 - 8r) + (20r - 32) = 0
Now, factor each group separately:
r(5r - 8) + 4(5r - 8) = 0
Notice that we have a common factor of (5r - 8) in both terms. Factor it out:
(5r - 8)(r + 4) = 0
Now, set each factor equal to zero and solve for r:
1. 5r - 8 = 0
5r = 8
r = 8/5
2. r + 4 = 0
r = -4
So, the solutions to the equation 5r^2 = 36r + 32 are:
r = 8/5 and r = -4.
Explanation: