1 x 31, 2 x 31/2 , 3 x 31/3 etc. 31 is a prime number.
1 x 93, 3 x 31, 31 x 3, 93 x 1.
Assuming you want two whole numbers that multiply to 279, the factors pairs are: 1 x 279 3 x 93 9 x 31 The prime factorisation of 279 is: 279 = 3^2 x 31 = 3 x 3 x 31
31/3 of 18 = 31/3 x 18 = 3x3+1/3 x 18/1 = 10/3 x 18/1 = 10x18/3x1 = 180/3 = 60
(75/3)+6 = X 25 + 6 = X 31 = X
x6 + 9= x6 - (-9) since i2 = -1= (x3)2 - 9i2 factor the difference of two squares= (x3 + 3i)(x3 - 3i) since 3 = (31/3)3 and -i = i3 we can write:= [x3 - (31/3)3i3] [x3 + (31/3)3i3]= [x3 - (31/3i)3] [x3 + (31/3i)3] factor the sum and the difference of two cubes= [(x - 31/3i)(x2 + 31/3ix + (31/3)2i2)] [(x + 31/3i)(x2 - 31/3ix + (31/3)2i2)]= [(x - 31/3i)(x2 + 31/3ix - (31/3)2)][(x + 31/3i)(x2 - 31/3ix - (31/3)2)]Thus, we have two factors (x - 31/3i) and (x + 31/3i),so let's find four othersAdd and subtract x2/4 to both trinomials[x2 - x2/4 + (x/2)2 + 31/3ix - (31/3)2] [x2 - x2/4 + (x/2)2 - 31/3ix - (32/3)2] combine and factor -1= {3x2/4 - [((x/2)i))2 - 31/3ix + (31/3)2]}{3x2/4 - [((x/2)i))2 + 31/3ix + (32/3)2]} write the difference of the two squares= {((3)1/2x/2))2 - [(x/2)i - 31/3]2}{((3)1/2x/2))2 - [(x/2)i + 32/3]2]} factor the difference of two squares= {[(31/2/2)x - ((1/2)i)x - 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - (((1/2)i)x + 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]}= {[(31/2/2)x - ((1/2)i)x + 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - ((1/2)i)x - 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]} simplify= {[((31/2 - i)/2))x + 31/3)] [((31/2 + i)/2))x - 31/3)]} {[((31/2 - i)/2))x - 31/3)] [((31/2+ i)/2))x + 31/3)]}so we have the 6 linear factors of x2 + 9.1) (x - 31/3i)2) (x + 31/3i)3) [((31/2 - i)/2))x + 31/3)]4) [((31/2 + i)/2))x - 31/3)]5) [((31/2 - i)/2))x - 31/3)]6) [((31/2+ i)/2))x + 31/3)]Check: Multiply:[(1)(2)][(3)(5)][(4)(6)]A) (x - 31/3i)(x + 31/3i) = x +(31/3)2B) [((31/2 - i)/2))x + 31/3)] [((31/2 - i)/2))x - 31/3)] = [(1 - (31/2)i)/2]x2 - (31/3)2C) [((31/2 + i)/2))x - 31/3)][((31/2+ i)/2))x + 31/3)] = [(1 + (31/2)i)/2]x2 - (31/3)2Multiply B) and C) and you'll get x4 - (31/3)2x2 + (31/3)4Now you have:[x +(31/3)2][x4 - (31/3)2x2 + (31/3)4] = x6 + 9
1 x 31, 2 x 31/2 , 3 x 31/3 etc. 31 is a prime number.
The factors of 372 are: 1 2 3 4 6 12 31 62 93 124 186 372Prime factors are 2, 2, 3, 31Others, being combinations of the primes, are:2 x 2 = 42 x 3 = 62 x 2 x 3 = 122 x 31 = 623 x 31 = 932 x 2 x 31 = 1242 x 3 x 31 = 1862)3722)186. 3)93. . 31so, combining these factors,1, 2, 3, 2*2, 2*3, 2*2*3, 31, 2*31, 3*31, 2*2*31, 2*3*31, 2*2*3*311, 2, 3, 4, 6, 12, 31, 62, 93, 124, 186, 372372186,293,2,231,3,2,2372186,293,2,231,3,2,21 x 3722 x 1863 x 1244 x 936 x 6212 x 312 x 2 x 932 x 3 x 622 x 6 x 313 x 4 x 312 x 2 x 3 x 31
837 = 3 x 3 x 3 x 31 or 33 x 31
1 x 93, 3 x 31, 31 x 3, 93 x 1.
10 + 3(x + 2) = 31
186 = 93 x 2 93 = 31 x 3 Since 2,3 and 31 are prime then 186 = 2 x 3 x 31
The factor pairs of 93 are 1 x 93 and 3 x 31
1 times 93 or 3 times 31
31 x 3It is: 3*31 = 93
6x - 2x - 3 = 31 grouping the expressions with x: 6x - 2x = 31 + 3 simplifying: 4x = 34 solving for x x = 34/4 = 8.5
3 x 3 x 3 x 3 = 81 3 x 31 = 93