8. 300 ends in 0 or 5, so 5 is a factor 300 is divisible by 4 and 5, so 20 is a factor 300 is even and 3 + 0 + 0 = 3, so 6 is a factor Leaving 8 to be the non factor Testing if 8 is a factor: 4 x 3 + 2 x 0 + 0 = 12; 4 x 0 + 2 x 1 + 2 = 4 which is not 8, so 300 is not divisible by 8.
x2 - 9x = -8 ∴ x2 - 9x + 8 = 0 ∴ (x - 1)(x - 8) = 0 You can then solve it as well: ∴ x ∈ {1, 8}
x2-9x+8=0 has two solutions:x = 8x = 1
It is when factored: (N-8)(N+3)
4x - 8 = 0 when x = 2 Substitute this value of x in 16x3 + 4x2 - 144 to give 16*8 + 4*4 - 144 = 128 + 16 - 144 = 144 - 144 = 0 So, by the remainder theorem, (x-2) is a factor of 16x3 + 4x2 - 144 Also, the coeffs are all divisibl by 4 so 4 is also a factor (x-2) is a factor, 4 is a factor so 4*(x-2) = (4x - 8) is a factor.
8. 300 ends in 0 or 5, so 5 is a factor 300 is divisible by 4 and 5, so 20 is a factor 300 is even and 3 + 0 + 0 = 3, so 6 is a factor Leaving 8 to be the non factor Testing if 8 is a factor: 4 x 3 + 2 x 0 + 0 = 12; 4 x 0 + 2 x 1 + 2 = 4 which is not 8, so 300 is not divisible by 8.
(x-8)(x+8) = x2-64
x2 - 9x = -8 ∴ x2 - 9x + 8 = 0 ∴ (x - 1)(x - 8) = 0 You can then solve it as well: ∴ x ∈ {1, 8}
No. Two is a factor in even numbers, which end in 0, 2, 4, 6, or 8.
Undefined: You cannot divide by zero
I think you mean r2 -7r -8 = 0. If so, then factor it: (r+1)(r-8) = 0 Then set r+1 = 0, r+1-1=-1, r = -1 Set r-8 =0, r-8+8 = 8 so r = 8
-(h - 4)(h + 8)
x2-9x+8=0 has two solutions:x = 8x = 1
It is when factored: (N-8)(N+3)
0
8 is a factor of 32 and of itself, so the answer is 8. Since 8 is a factor of 32, it is automatically the GCF.Since 8 is a factor of 32, it is automatically the GCF.It is 8
Any non-zero integer can be a factor. So 8 and 13.