Dayaan M. answered • 08/26/26
Earned A’s Twice in Precalculus | 5 Years of Tutoring Experience
In order to build a polynomial from a list of conditions like this, the trick is to treat each condition as information about one specific part of the function and then assemble them one at a time. None of it has to be solved all at once.
Start with the zeros, since those give us the factors. If r is a zero, then (x - r) is a factor, and if that zero has multiplicity 3, the factor appears three times. We are told 2 is a zero of multiplicity 3, so we begin with (x - 2)^3.
Now look at the degree. We need degree 6, and (x - 2)^3 only uses up 3 of that, so three factors are still missing. We also need four real zeros that are not all distinct, and 2 is one of them, so we need three more zeros, each appearing once. That fits perfectly, three more factors and 3 + 3 = 6.
Next, saying f behaves like y = 2x^6 for large values of the absolute value of x is just telling us the leading coefficient is 2. When x gets very large, only the highest power really matters, so we put a 2 out front.
So far we have:
f(x) = 2(x - 2)^3(x - a)(x - b)(x - c)
and the last condition, f(0) = 8, is what pins down a, b and c. Plugging in 0:
f(0) = 2(-2)^3(-a)(-b)(-c) = 2(-8)(-abc) = 16abc
Setting that equal to 8 gives abc = 1/2. Remember, the problem says there is more than one possible answer, so we just need any three numbers that are different from each other and from 2, whose product is 1/2. Let us take 1, -1 and -1/2, since (1)(-1)(-1/2) = 1/2.
f(x) = 2(x - 2)^3(x - 1)(x + 1)(x + 1/2)
Let us check it against every condition. The degree is 3 + 1 + 1 + 1 = 6. The distinct real zeros are 2, 1, -1 and -1/2, which is four, and they are not all distinct in the sense that 2 repeats three times. The leading coefficient is 2. And f(0) = 2(-8)(-1)(1)(1/2) = 8.
So, our final answer is f(x) = 2(x - 2)^3(x - 1)(x + 1)(x + 1/2), and any other three numbers multiplying to 1/2 would work just as well.
