when is the balloon exactly 980m above the ground
When h(t)=980, -2t^3+3t^2+149t-570=0. The factors of 570=2*3*5*19. From these we can make several composite numbers that could be factors if it is assumed that there are rational zeroes. We need three zeroes or roots and the possible factors are (2,3,95), (2,15,19), (6,5,19), (3,10,19), (2,5,57), (3,5,38). The factors of 2 (t^3 coefficient) are (1,2) only. By testing each factor by dividing the cubic expression one at a time by the factors, both positive and negative values, using, say, synthetic division, we can reduce the cubic to a quadratic. When we do this, we discover that 5 is a zero, and when we divide the expression by t-5 we get -2t^2-7t-114 which further factorises into -(t-6)(2t+19). So the zeroes are 5, 6, -19/2. We can reject the last one, because it's negative and out of the given range (0 to 10).
Therefore t=5 or 6 seconds are valid solutions. The balloon reaches 980m at 5 seconds then continues to rise then drops to 980m at 6 seconds.
Check: -250+75+745+410=980=-432+108+894+410. Read More: ...