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ki77a [65]
2 years ago
6

Does anyone know how to do this (algebra 2)

Mathematics
1 answer:
Ksivusya [100]2 years ago
5 0
100 < 50(1.12)t
2 < (1.12)t
ln 2 < t ln (1.12)
0.6931 < t(0.1133)
6.117 < t
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Describe the steps you would use to factor 2x3 + 5x2 – 8x – 20 completely. Then state the factored form.
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The polynomial 2x^3 + 5x^2-8x-20 may have solutions which are the divisors of -20, therefore -20 has the following divisors: \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.
If x=1, then 2\cdot1^3 + 5\cdot1^2-8\cdot1-20=-20\neq 0,
if x=-1, then 2\cdot(-1)^3 + 5\cdot(-1)^2-8\cdot(-1)-20=-9\neq 0,
if x=2, then 2\cdot2^3 + 5\cdot2^2-8\cdot2-20=0, then x=2 is a solution and you have the first factor (x-2). 
If x=-2, then 2\cdot(-2)^3 + 5\cdot(-2)^2-8\cdot(-2)-20=0, then x=-2 is a solution, so you have the second factor (x+2).
Since x-2 and x+2 are two factors of 2x^3 + 5x^2-8x-20 , then the polynomial x^2-4 is a divisor of 2x^3 + 5x^2-8x-20 and dividing the polynomial 2x^3 + 5x^2-8x-20 by x^2-4 you obtain 
 2x^3 + 5x^2-8x-20=(x-2)(x+2)(2x+5).











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