Find a polynomial f(x) of degree 4 that has the following zeros. 0, -5, 8, -2

Mathematics

1 answer:

Roman55 [17]3 years ago

3 0

Answer:

The polynomial of the function degree '4'

x⁴ - x³ -46 x² -80x =0

Step-by-step explanation:

Explanation:-

Given zeros are 0 ,-5,8,-2

x =0 , x=-5 ,x=8 and x=-2

(x-0) , ( x+5) , (x-8) and (x+2)

The polynomial function with degree '4'

(x-0)(x-(-5)(x-8)(x-(-2) =0

x ( x+5) (x-8) (x+2) =0

(x² +5x)(x-8)(x+2) =0

(x³ -8 x²+5x² -40 x)(x+2) =0

( x³ -3 x² - 40 x) (x +2) =0

x⁴ + 2 x³ - 3 x³- 6 x²-40 x²- 80 x =0

x⁴ - x³ -46 x² -80x =0

Final answer:-

The polynomial of the function degree '4'

x⁴ - x³ -46 x² -80x =0

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PLEASE HELP the formula m=12,000+12,000rt/12t gives keri's monthly loan payment where t is the annual interest rate and t is the

tatiyna

Answer:

The answer is below

Step-by-step explanation:

The formula m = (12,000 + 12,000rt)/12t gives Keri's monthly loan payment, where r is the annual interest rate and t is the length of the loan, in years. Keri decides that she can afford, at most, a $275 monthly car payment. Give an example of an interest rate greater than 0% and a loan length that would result in a car payment Keri could afford. Provide support for your answer.

Answer: Let us assume an annual interest rate (r) = 10% = 0.1. The maximum monthly payment (m) Keri can afford is $275. i.e. m ≤ $275. Using the monthly loan payment formula, we can calculate a loan length that would result in a car payment Keri could afford.

$m=\frac{12000+12000rt}{12t}\\ but\ m\leq275, \ and \ r=10\%=0.1\\275= \frac{12000+12000(0.1)t}{12t}\\275= \frac{12000}{12t} +\frac{12000(0.1)}{12t}\\275= \frac{1000}{t} + 100\\275-100= \frac{1000}{t} \\175= \frac{1000}{t} \\175t = 1000\\t= \frac{1000}{175}\\ t=5.72\ years$