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Determine algebraically where the polynomial function that has zeros at 2,3 and -5 and passes through the point (4,36) has a value of 120?
2 Answers
- JLv 75 years ago
k(x-2)(x-3)(x+5) has value 36 at x= 4, plug in and get
18k = 36, so k=2 and the polynomial is
2(x-2)(x-3)(x+5) which multiplies out to
2x^3 - 38x + 60 so we have to solve
2x^3 - 38x + 60 = 120 for x
2x^3 - 38x - 60 = 0 factor
2(x-5)(x+3)(x+2) = 0 so
x=5, or x=-3, or x=-2
interesting
- ?Lv 75 years ago
f(x) = a (x - 2) (x - 3) (x + 5)
f(4) = 36
a (4 - 2) (4 - 3) (4 + 5) = 36
18a = 36
a = 2
f(x) = 2 (x - 2) (x - 3) (x + 5)
Now we find where function has value of 120
2 (x - 2) (x - 3) (x + 5) = 120
(x - 2) (x - 3) (x + 5) = 60
x^3 - 19x + 30 = 60
x^3 - 19x - 30 = 0
(x + 2) (x + 3) (x - 5) = 0
x = -2, -3, 5
