Solving the Equation (x+2)(x5)(x6)(x+1)=144
This equation presents a challenge as it's a quartic equation (an equation with the highest power of x being 4). Here's how we can solve it:
1. Expand the Equation
First, let's expand the lefthand side of the equation:

Step 1: Expand (x+2)(x5) and (x6)(x+1) (x+2)(x5) = x²  3x  10 (x6)(x+1) = x²  5x  6

Step 2: Expand the entire equation (x²  3x  10)(x²  5x  6) = 144 x⁴  8x³  11x² + 90x + 60 = 144

Step 3: Move all terms to one side x⁴  8x³  11x² + 90x  84 = 0
2. Finding Possible Rational Roots
Now we have a quartic equation. We can use the Rational Root Theorem to find possible rational roots. The theorem states that any rational root of the equation must be of the form p/q, where:
 p is a factor of the constant term (84)
 q is a factor of the leading coefficient (1)
The factors of 84 are ±1, ±2, ±3, ±4, ±6, ±7, ±12, ±14, ±21, ±28, ±42, ±84. The factors of 1 are ±1.
Therefore, the possible rational roots are: ±1, ±2, ±3, ±4, ±6, ±7, ±12, ±14, ±21, ±28, ±42, ±84.
3. Testing the Possible Roots
We can test each of these possible roots by substituting them into the equation. We find that x = 7 is a root.
4. Factoring the Equation
Since x=7 is a root, (x7) must be a factor of the equation. We can use polynomial long division or synthetic division to divide the quartic equation by (x7):
x³  x²  18x + 12
x7  x⁴  8x³  11x² + 90x  84
x⁴  7x³

x³  11x²
x³ + 7x²

18x² + 90x
18x² + 126x

36x  84
36x + 252

336
This gives us: (x7)(x³  x²  18x + 12) = 0
5. Finding Remaining Roots
Now we have a cubic equation: x³  x²  18x + 12 = 0. Finding the roots of this cubic equation is a bit more complex. You can use numerical methods like NewtonRaphson or graphing calculators to find the approximate solutions.
The solutions to the cubic equation are approximately:
 x ≈ 4.28
 x ≈ 1.76
 x ≈ 4.52
6. Summary of Solutions
Therefore, the solutions to the original equation (x+2)(x5)(x6)(x+1)=144 are:
 x = 7
 x ≈ 4.28
 x ≈ 1.76
 x ≈ 4.52