Degree 4; zeros: -5+5i; 4 multiplicity 2? Enter the polynomial. f(x)=a(.........)

Question

Degree 4; zeros: -5+5i; 4 multiplicity 2?
Enter the polynomial.
f(x)=a(.........)

Answer 1

Here are our zeros:

 

x = -5 + 5i

x = -5 - 5i

 

These two zeros are conjugates.  They were obtained from using the quadratic formula. 

 

The other root is x=4, which appears twice.

 

 

Our polynomial will then be

 

f(x) = [x + (5 - 5i)][x + (5 + 5i)](x - 4)(x - 4)

 

f(x) = [x² + (5 + 5i)x + (5 - 5i)x + (25 + 25)] (x² - 8x + 16)

 

f(x) = (x² + 5x + 5ix + 5x - 5ix + 50)(x² - 8x + 16)

 

f(x) = (x² + 10x + 50)(x² - 8x + 16)

 

f(x) = x²(x² - 8x + 16) + 10x(x² - 8x + 16) + 50(x² - 8x + 16)

 

f(x) = x⁴ - 8x³ + 16x² + 10x³ - 80x² + 160x + 50x² - 400x + 800

 

f(x) = x⁴ + 2x³ - 14x² - 240x + 800

Answer 2

Assuming that the polynomial has real coefficients, complex roots occur in conjugate pairs.  So, since -5+5i is a zero of f(x), so is -5-5i.  

 

Since degree (f) = 4, there are at most 4 distinct zeros.  Since 4 is a zero with multiplicity two, the zero, 4, is counted as two of the zeros.

 

So, the zeros of f(x) are 4, 4, -5+5i, -5-5i.

 

f(x) = a(x-4)(x-4)[x-(-5+5i)][x-(-5+5i)]

 

      = a(x²-8x+16)[(x+5)-5i][(x+5)+5i]

 

      = a(x²-8x+16)[(x+5)² - (5i)²]

 

      = a(x²-8x+16)(x²+10x+25-25i²)

 

      = a(x²-8x+16)(x²+10x+50)

 

      = a(x⁴+10x³+50x²-8x³-80x²-400x+16x²+160x+800)

 

      = a(x⁴+2x³-14x²-240x+800)

 

NOTE:  a is the leading coefficient.  a is a nonzero number.  To determine the value of a, further info is needed.