If P(x)= 3x^4 - 5x^3 - 17x^2 + 13x + 6, and P(1)=0, and P(-2)=0, then the factorization of P(x) is?
1 answer:
Answer:
P(x) = (x - 1)(x + 2)(x - 3)(3x + 1)
Step-by-step explanation:
Since P(1) = 0 and P(- 2) = 0, then
(x - 1) and (x + 2) are factors of P(x)
(x - 1)(x + 2) = x² + x - 2 ← is also a factor of P(x)
dividing 3 - 5x³ - 17x² + 13x + 6 by x² + x - 2 gives
P(x) = (x - 1)(x + 2)(3x² - 8x - 3) = (x - 1)(x + 2)(x - 3)(3x + 1)
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Answer:
<h3><u>Option 1</u></h3>Earn $50 every month.
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000
This is a <u>linear function</u>.
<h3><u>Option 2</u></h3>Earn 3% interest each month.
(Assuming the interest earned each month is <u>compounding interest</u>.)
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000
This is an <u>exponential function</u>.
<h3><u>Table of values</u></h3><u />
From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.
However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the <u>point of intersection</u>.
From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.
<h3><u>Conclusion</u></h3>If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.
