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3 years ago

What is the polynomial function of least degree whose only zeros are -3, ,-2, and 4 ?

Mathematics

1 answer:

victus00 [196]3 years ago

7 0

Answer: (x³ - 2x² - 11x + 12)

Step-by-step explanation:

Recall that zeroes can be transformed into factors by subtracting them from x. This gives us the following factors:

(x - 1)(x + 3)(x - 4)

Now, if you multiply the first two factors together, you get the following:

(x² + 2x - 3)

Multiply that by the last factor, (x - 4), and you get this:

(x³ + 2x² - 3x - 4x² - 8x + 12)

This can be simplified:

(x³ - 2x² - 11x + 12)

And there's your final answer. Hope this helped! (x³ - 2x² - 11x + 12)

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A corporate bond has a coupon rate of 5.5 percent, a $1,000 face value, and matures three years from today. The corporation is i

melomori [17]

Answer:

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

Step-by-step explanation:

= (probability of entire face value paid*face value+probability of entire face value not paid*percent of face value paid*face value)/(1+discount rate)^years to maturity

probability of entire face value paid = 75%

face value = 1000

probability of entire face value not paid = 25%

percent of face value paid= 60%

discount rate = 15%

years to maturity = 3

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

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