1. What are the least, and most, number of distinct real roots of a 6th degree polynomial?

Question

1. What are the least, and most, number of distinct real roots of a 6th degree polynomial?
2. What is i103?
3. The zeroes of f(x) = x² – 8x + 17 are?
4. If f(x) = x⁸ - 1 is divided by x -2, the remainder would be?
5. What are the possible rational roots of f(x) = 5x⁴ - 173x³ -16x² -7x -15, according to the rational root theorem? “+-“ means “plus or minus”.

Answer 1

1. A polynomial can't have more roots than the degree.  So, a sixth degree polynomial, has at most 6 distinct real roots.  For example,        (x-1)(x-2)(x-3)(x-4)(x-5)(x-6) has degree 6 and has 6 distinct real roots.  

A polynomial may have no real roots.  So, the fewest number of real roots of a polynomial with degree 6 could be 0.  This would be the case if the graph of y = polynomial has no x-intercepts. For example, x⁶+1 has degree 6 and has no real roots [and the graph of

f(x) = x⁶+1 has no x-intercepts].     

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2. To find i raised to a positive integer power, divide the power by 4 and use the remainder as the power.  So, if we divide 103 by 4, the remainder is 3.  Therefore, i¹⁰³ = i³ = -i.

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3.  x²-8x+17 = 0     x = [8 ± √(-4)]/2 = [8 ± 2i]/2 = 4±i

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4.  According to the Remainder Theorem, if a polynomial f(x) with real coefficients is divided by x - c, then the remainder is f(c).  

 

So, if f(x) = x⁸ - 1, then the remainder when f(x) is divided by x-2 is f(2) = 2⁸ - 1 = 255.  

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5. If f(x) has a rational root, p/q, then p is a divisor of the constant term and q is a divisor of the leading coefficient.  

So, p = ±(1, 3, 5, 15) and q = ±(1, 5)

Therefore, p/q = ±(1, 1/5, 3, 3/5, 5, 15)

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Answer 2

2)

 

We know that    i² = -1

 

 

If we raise both sides of the equation by 51,

 

(i²)⁵¹ = (-1)⁵¹

 

i¹⁰² = -1

 

 

Now we multiply both sides of the resulting equation by i.

 

(i¹⁰²) * i = (-1) * i
 
i¹⁰³ = -i

 

 

3)

 

Set f(x) equal to zero and solve for x.

 

0 = x² - 8x + 17

 

 

You will need to use the quadratic formula to solve this one.

 

 

4)

 

Use long division.

 

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(x - 2) | x⁸ + 0x⁷ + 0x⁶ + 0x⁵  0x⁴ + 0x³ + 0x² + 0x - 1

 

 

 

You can also use synthetic division, if you are familiar with that approach.

 

 

5)

 

The possible roots are

 

±1, ±3 , ±5 , ±15 , ±1/5 , and ±3/5