form a fifth-degree polynomial function with real coefficients such that 5i, 1-3i, and -3 are zeroes and f(0)=1500

Question

 help!

Answer 1

A fifth degree polynomial must have 5 roots.  All Complex roots must come in pairs, so if 5i is root then -5i is also a root.  If 1-3i is a root, then 1+3i must also be a root.

Turn the roots into factors and multiply:

 

(x - 5i)(x + 5i)(x-1-3i)(x-1+3i)(x+3)

Answer 2

Your complex zeros have conjugates.  So if you include them into the count, you will have a total of 5 zeros.

 

x₁ = -3

x₂ = 5i

x₃ = -5i

x₄ = 1 - 3i

x₅ = 1 + 3i

 

 

Next, use these zeros to put them in factored form.

 

f(x) = C(x + 3)(x - 5i)(x + 5i)(x - [1 + 3i])(x - [1 - 3i])

 

where C is a constant.  We put this for the y-intercept condition.

 

 

Now, we just expand to get the polynomial without imaginary terms.

 

f(x) = C(x + 3)(x² + 25)(x² - x(1 - 3i) - x(1 + 3i) + (1 + 9))

 

f(x) = C(x + 3)(x² + 25)(x² - 2x + 10)

 

f(x) = C(x³ + 3x² + 25x + 75)(x² - 2x + 10) 

 

 

Using the fact that f(0)=1500, we can solve for C.

 

1500 = 75C (10)

 

1500 = 750C

 

2 = C

 

 

f(x) = 2(x³ + 3x² + 25x + 75)(x² - 2x + 10)

 

f(x) = (2x³ + 6x² + 50x + 150)(x² - 2x + 10)

 

f(x) = 2x³(x² - 2x + 10) + 6x²(x² - 2x + 10) + 50x(x² - 2x + 10) + 150(x² - 2x + 10)

 

 

Now just simplify by distributing the terms to get the function in standard form.

Answer 3

Hi Jessica.

 

To get the polynomial you are looking for there are a few pieces of information that we need to add.

 

ALL COMPLEX ROOTS COME IN PAIRS.  So they have purposefully left out two roots to see if you know that you must find the "other" root for each complex number.

 

The two roots are complex conjugates.  That means that if one is "5i", the other is "-5i".  If you is "1-3i", the other is "1+3i"

 

Now you have 5 roots and can make a fifth degree polynomial.

 

 

The roots are:

 

x=5i,  x= -5i,    x=1-3i,     x=1+3i,   x= -3

 

 

For each "x=", move everything to the left side so that these become factors instead of zeros.

 

x-5i=0,    x+5i=0,   x-1+3i = 0,   x-1-3i = 0, x+3=0

 

Now your factors are:

 

(x-5i)(x+5i)(x-1-3i)(x-1+3i)(x+3)

 

 

 

Multiply the first two and get

 

(x-5i)(x+5i) = x² +(5x)i - (5x)i - 25i²

 

The middle terms cancel each other

25i² = (25)(-1) = -25

 

So this becomes (x² - (-25)) = x²+25

 

 

The 3rd and 4th terms multiplied together give:

 

(x-1+3i)(x-1-3i)=

 

x² - x - (3x)i - x + 1 +3i +(3x)i - 3i - 9i²

 

Notice that all of the "i" terms go away.   

 

9i² = (9)(-1) = -9

 

The -(3x)i cancels with the (3x)i

The -3i  cancels with the 3i

 

All that are left are real terms.

 

x² - 2x -(-9)

 

x² - 2x + 9

 

 

Now you have all real coefficients to work with:

 

 

Your original :  (x-5i)(x+5i)(x-1-3i)(x-1+3i)(x+3)

 

Has become:  (x² + 25)(x² -2x + 9)(x+3)

 

Multiply the first two.   Then use that result to multiply with the remaining factor (x+3)

 

 

 

(x²+25)(x² - 2x + 9) = 

 

x⁴ - 2x³ + 9x² + 25x² - 50x + 225

 

 

This simplifies to:

 

x⁴ - 2x³ + 34x² - 50x + 225

 

 

Last step is to multiply this by the (x+3) and you will get a 5th-degree polynomial

 

 

(x+3)(x⁴ - 2x³ + 34x² - 50x + 225) =

 

x⁵ -2x⁴ +34x³ - 50x² + 225x + 3x⁴ -6x³ +102x² - 150x +675

 

 

Combine like terms to get your final answer, a 5th-degree polynomial with all real coefficients:

 

x⁵ + x⁴ +28x³ + 52x² +75x +675