784
2 x 2 x 7 x 11 or 2^2 x 7^1 x 11^1
f(x) = 2x + 1 g(x) = x^2 - 7 So f*g(x) = f(g(x)) = f(x^2 - 7) = 2*(x^2 - 7) + 1 = 2*x^2 - 14 + 1 = 2*x^2 - 13
1 x 14, 2 x 7, 7 x 2, 14 x 1
10 + 2(x - 1) + 7 x 3x + 1 =10 + 2x - 2 + (7 x 3x) + 1 =10 + 2x - 2 + 21x + 1 =2x + 21x + 10 - 2 + 1 =23x + 9
2744 = 1 x 1 x 2 x 2 x 2 x 7 x 7 x 7 Make 3 groups of these factors and multiply the numbers in each group to get 3 numbers, eg: {1}, {1}, {2, 2, 2, 7, 7, 7} = 1 x 1 x 2744 {1}, {1, 2}, {2, 2, 7, 7, 7} = 1 x 2 x 1372 {1, 2}, {1, 2, 7}, {2, 7, 7} = 2 x 14 x 98 {1, 2, 7}, {1, 2, 7}, {2, 7} = 14 x 14 x 14 There are many more such answers.
784
70 = 7 x 10 = 7 x 2 x 5 56 = 7 x 8 = 7 x 2 x 2 x 2 Common factors are 1, 2, 7 and 14
196^(1/2) = 14 or 196^(0.5) = 14 Since 196 is an even number it will divide by '2' Hence 2)196 2)98 7)49 7)7 = 1 Reduce to '1' Hence 2 x 2 x 7 x 7 = 2^(2) x 7^(2) = (2 x 7)^(2) square rooting both numbers. [(2 x 7)^(2)]^(1/2) = [2 x 7]^(2/2) = [2 x 7 ] ^(1) = (2 x 7) = 14
This does readily factor!!! So by trial and error. Discount '0' because the you are left with '-7' Try '-1 (-1)^(3) - 5(-1)^(2) - 13(-1) - 7 => -1 - 5 + 13 - 7 adding = '0' Hence x = -1 Therefore ( x + 1) is a factor. (x+1) ( x^(3) - 5(x^(2) - 13x - 7 = (x^(2) -x^(3) -x^(2) = -6x^(2) - 13x = ( x^(2) -6x 6x^(2) + 6x = -7(x + 1) ( x^(2) - 6x + 7) (x + 1)(x^(2) - 6x + 7) is fully factored. If you factor the Quadratic, you will move into the region of IMGINARY (i) numbers.
2 x 2 x 7 x 11 or 2^2 x 7^1 x 11^1
f(x) = 2x + 1 g(x) = x^2 - 7 So f*g(x) = f(g(x)) = f(x^2 - 7) = 2*(x^2 - 7) + 1 = 2*x^2 - 14 + 1 = 2*x^2 - 13
Factors of 98 are 1, 2, 7, 14, 49, and 98.
Factor them. 2 x 2 1 x 7 2 x 2 x 3 3 x 7 The answer is 1 x 2 x 2 x 3 x 7 or 84
The power is 1. 2 x 7 x 11 = 21 x 71 x 111 or (2 x 7 x 11)1
1 x 98, 2 x 49, 7 x 14, 14 x 7, 49 x 2, 98 x 1 = 98
1 x 14, 2 x 7, 7 x 2, 14 x 1