Find an nth-degree.

Question

n=4

i and 3i are zeros

f(-2)=130

Answer 1

Every complex number has a partner complex number.

So, if i is a solution, then -i is also a solution.

If 3i is a solution, then -3i is a solution.

x= i x=-i x=3i x=-3i

(x-i)(x+i)(x-3i)(x+3i)=0 Now foil

(x²-i²)(x²-9i²) i^=-1

(x²-(-1))(x²-9(-1))

(x²+1)(x²+9) Now foil again

x⁴+9x²+x²+9

f(x)= x⁴+10x²+9

You can check if f(-2)=130

(-2)⁴+10(-2)²+9

16+10(4)+9

16+40+9

f65

f(-2)=65 which is half of what we needed.

That is our clue that a GCF was taken out before solving.

Instead of f(x)=x⁴+10x²+9, f(x)=2(x⁴+10x²+9) or 2x⁴+20x²+18

You can check your answer by trying to solve this back out!

2x⁴+20x²+18=0 Factor out a GCF

2(x⁴+10x²+9)=0 Divide by GCF

x⁴+10x²+9=0 Now Factor. What multiples to get 9, adds to get 10

(x²+1)(x²+9)=0 Set each factor=0

x²=-1 x²=-9 Square root both sides (don't forget your +/-)

x=+/- i x=+/- 3i

Which is what we started with!

Hope this helps!

Answer 2

f(x) = a(x - i)x + i)(x - 3i)(x + 3i

f(x) = a(x² + 1)(x² + 9)

f(x) = a(x⁴ + 10x² + 9)

f(-2) = 130

130 = a(16 + 40 + 9)

130 = a(65)

2 = a

f(x) = 2x⁴ + 20x² + 18