2*(4! + 3) + 1 = 2*(24 + 3) + 1 = 2*27+1 = 54+1 = 55
2x4 - x3 - 21x2 + 9x + 27 = 0 Let f(x) = 2x4 - x3 - 21x2 + 9x + 27. All possible rational zeros = (factors of the constant term, 27)/(factors of the leading coefficient, 2) = ±1, ±3, ±9, ±27)/(±1, ±2) = ±1, ±3, ±9, ±27, ±1/2, ±3/2, ±9/2, ±27/2. Using synthetic division, test all possible rational zeros: -1] 2 -1 -21 9 27 (2)(-1) = -2; -1 + -2 = -3 (-3)(-1) = 3; -21 + 3 = -18 (-18)(-1) = 18; 9 + 18 = 27 (27)(-1) = -27; 27 + -27 = 0 (the zero remainder shows that -1 is a zero) -3] 2 -1 -21 9 27 (2)(-3) = -6; -1 + -6 = -7 (-7)(-3) = 21; -21 + 21 = 0 (0)(-3) = 0; 9 + 0 = 9 (9)(-3) = -27; 27 + -27 = 0 (the zero remainder shows that -3 is a zero) 3/2] 2 -1 -21 9 27 (2)(3/2) = 3; -1 + 3 = 2 (2)(3/2) = 3; -21 + 3 = -18 (-18)(3/2) = -27; 9 + -27 = -18 (-18)(3/2) = -27; 27 + -27 = 0 (the zero remainder shows that 3/2 is a zero) 3] 2 -1 -21 9 27 (2)(3) = 6; -1 + 6 = 5 (5)(3) = 15; -21 + 15 = -6 (-6)(3) = -18; 9 + -18 = -9 (-9)(3) = -27; 27 + -27 = 0 (the zero remainder shows that 3 is a zero) Thus, the rational zeros are -3, -1, 3/2, and 3.
27 = 27/1 = (2*27)/2 = (3*27)/3 = ...
Assuming the order of the numbers is not important, then ((4 x 2) + 1) x 3 = 27.
0.1111
3*(2*4 + 1) = 3*(8 + 1) = 3*9 = 27
2*(4! + 3) + 1 = 2*(24 + 3) + 1 = 2*27+1 = 54+1 = 55
2x4 - x3 - 21x2 + 9x + 27 = 0 Let f(x) = 2x4 - x3 - 21x2 + 9x + 27. All possible rational zeros = (factors of the constant term, 27)/(factors of the leading coefficient, 2) = ±1, ±3, ±9, ±27)/(±1, ±2) = ±1, ±3, ±9, ±27, ±1/2, ±3/2, ±9/2, ±27/2. Using synthetic division, test all possible rational zeros: -1] 2 -1 -21 9 27 (2)(-1) = -2; -1 + -2 = -3 (-3)(-1) = 3; -21 + 3 = -18 (-18)(-1) = 18; 9 + 18 = 27 (27)(-1) = -27; 27 + -27 = 0 (the zero remainder shows that -1 is a zero) -3] 2 -1 -21 9 27 (2)(-3) = -6; -1 + -6 = -7 (-7)(-3) = 21; -21 + 21 = 0 (0)(-3) = 0; 9 + 0 = 9 (9)(-3) = -27; 27 + -27 = 0 (the zero remainder shows that -3 is a zero) 3/2] 2 -1 -21 9 27 (2)(3/2) = 3; -1 + 3 = 2 (2)(3/2) = 3; -21 + 3 = -18 (-18)(3/2) = -27; 9 + -27 = -18 (-18)(3/2) = -27; 27 + -27 = 0 (the zero remainder shows that 3/2 is a zero) 3] 2 -1 -21 9 27 (2)(3) = 6; -1 + 6 = 5 (5)(3) = 15; -21 + 15 = -6 (-6)(3) = -18; 9 + -18 = -9 (-9)(3) = -27; 27 + -27 = 0 (the zero remainder shows that 3 is a zero) Thus, the rational zeros are -3, -1, 3/2, and 3.
33 = 27 with repetition, 3! = 3*2*1 = 6 without repetition.
(3 cubed +1 squared) - 2 = 26 also written as (3^3 + 1^2) - 2 = 26 or (27+1)-2 = 26 so 3,3,2,2 and 1 can make 26
32, 54, 80 / 2 = 16, 27, 40 16, 27, 40 / 2 = 8, 27, 20 8, 27, 20 / 2 = 4, 27, 10 4, 27, 10 / 2 = 2, 27, 5 2, 27, 5 / 2 = 1, 27, 5 1, 27, 5 / 3 = 1, 9, 5 1, 9, 5 / 3 = 1, 3, 5 1, 3, 5 / 3 = 1, 1, 5 1, 1, 5 / 5 = 1, 1, 1 SCM = 2^5 x 3^3 x 5 = 4,320
(y^2 + 3)(y^2 - 3y + 3)(y^2 + 3y + 3)
27 = 27/1 = (2*27)/2 = (3*27)/3 = ...
The divisors of 27 are: 1, 3, 9, 27.
1/9 x 27/1 = 27/9 = 3
The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54. The factor pairs of 54 are 1 x 54, 2 x 27, 3 x 18, and 6 x 9The factors of 54 are: 1 2 3 6 9 18 27 54.1, 2, 3, 6, 9, 18, 27, 54The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, & 541, 2, 3, 6, 54, 27, 18, 9
The factors of 18 are 1, 2, 3, 6, 9, and 18 The factors of 27 are 1, 3, 9, and 27