Degree 3; zeros: 3, 4 - i

jernplate8

jernplate8

Answered question

2021-06-01

Degree 3; zeros: 3, 4i

Answer & Explanation

Brittany Patton

Brittany Patton

Skilled2021-06-02Added 100 answers

Complex zeros come in conjugate pairs so we know that 4−i is the third zero given that 4+i is a zero.
If aa is a zero of a polynomial function, then xa is one of its factor. Using 2, 1+i, and 1i,
f(x)=a(x3)[x(4i)][x(4+i)]
By expanding,
f(x)=a(x3)[x2(4+i)x(4i)x+(4i)(4+i)]
f(x)=a(x3)[x24xix4x+ix+(16+1)]
f(x)=a(x3)(x28x+17)
f(x)=a[x2(x3)8x(x3)+17(x3)]
f(x)=a(x33x28x2+24x+17x51)
f(x)=a(x311x2+41x51)
For simplicity, let a=1:
f(x)=x311x2+41x51

2021-11-01

Degree 3: zeros: 1 + i and -5

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