No it is not.
Chat with our AI personalities
8=2*2*2*1 15=5*3*1 the only whole number that both are divisibl by is 1.
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.
1+8+5+4+5+5-5-5-5-5-5-5-5-5-5 = -17
50 50 25 25 25 25 20 x .05 50 25 25 50 25 5 5 5 5 5 50 5 5 5 5 5 5 5 5 5 5 25 25 25 5 5 5 5 5 25 25 5 5 5 5 5 5 5 5 5 5 25 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
0: (5x5)-(5x5) 1: (5:5)+5-5 2: (5:5)+(5:5) 3: (5+5+5):5 4: [(5x5)-5]:5 5: [(5-5)x5]+5 6: [(5x5)+5]:5 7: [(5+5):5]+5 10: 5+5+5-5 11: (5:5)+5+5 15: (5x5)-5-5 20: 5+5+5+5 24: (5x5)-(5:5) 25: (5x5)+5-5 26: (5x5)+(5:5) 30: [(5:5)+5]x5 35: (5x5)+5+5 45: [(5+5)x5]-5 50: (5x5)+(5x5) 55: [(5+5)x5]+5 75: (5+5+5)x5 120: (5x5x5)-5 130: (5x5x5)+5 625: 5x5x5x5