Q: How do you factor the monomial 24z to the second power?

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24(w + z)

Yes, of multiplication.

24z+4 or 6(4z+1)

A polynomial has 3 or more terms. It can have exponents and variables. Example: 24x + 74y - 24z 24x = term 1 74y = term 2 24z = term 3 EDIT: Polynomials can have 1 or 2 terms as well. But those are special, they are monomials and binomials.

Both Exhaust & Inlet Clearances are 0.30mm = 0.012"...or..12 thou HOT

well actually how you beat the rock at the game is like say I'm the rock and I'm at the game its like multiplying 24z=5/7 a pint of silver

Given: 3x + y + 2z = 1 2x - y + z = -3 x + y - 4z = -3 Take any one of the equations (we'll use the first one), and solve for any one of the variables (we'll use y): y = 1 - 2z - 3x Now plug that value of y into the latter two equations: 2x - (1 - 2z - 3x) + z = -3 x + (1 - 2z - 3x) - 4z = -3 Now take either of those (again, we'll use the first one), and solve it for either of the remaining variables (we'll go for x): 2x - 1 + 2z + 3x + z = -3 ∴ 5x = -2 - 3z ∴ x = (3z + 2) / -5 Now take that value, and plug it into our other equation that uses x and z: (3z + 2) / -5 + 1 - 2z - 3(3z + 2) / -5 - 4z = -3 Then solve for z: ∴ 2(3z + 2) / 5 - 6z = -4 ∴ (6z + 4 - 30z) / 5 = -4 ∴ 4 - 24z = -20 ∴ 24z = 24 ∴ z = 1 Now we can take that value for z, and plug it back into our previous equation for x and z: x = (3z + 2) / -5 ∴ x = (3 + 2) / -5 ∴ x = -1 Finally, we can take those two values, and plug them into our equation for y: y = 1 - 2z - 3x ∴ y = 1 - 2 + 3 ∴ y = 2 So x = -1, y = 2, and z = 1 You can test these values by plugging them into each of the original three equations, and seeing if they solve correctly: 3x + y + 2z = 1 ∴ 3(-1) + 2 + 2(1) = 1 ∴ 2 + 2 - 3 = 1 ∴ 1 = 1 2x - y + z = -3 ∴ 2(-1) - 2 + 1 = -3 ∴ -2 - 2 + 1 = -3 ∴ -3 = -3 x + y - 4z = -3 ∴ -1 + 2 - 4 = -3 ∴ -3 = -3 which shows our answer to be correct.