p(-1+2i)=0
p(-1-2i)=0
p(2)=0
Step-by-step explanation:
Given polynomial is:
[tex]p(x) = (x-2)(x+1-2i)(x+1+2i)[/tex]
we have to find all the values on given values to find the correct options
[tex]p(-1+2i) = (-1+2i-2)(-1+2i+1-2i)(-1+2i+1+2i)\\=(-3+2i)(0)(4i)\\=0[/tex]
[tex]p(1-2i) = (1-2i-2)(1-2i+1-2i)(1-2i+1+2i)\\=(-1-2i)(-4i)(2)\\=(-1-2i)(-8i)\\= -8i-16i^2\\= -8i-16(-1)\\=-8i+16[/tex]
[tex]p(2i) = (2i-2)(2i+1-2i)(2i+1+2i)\\=(2i-2)(1)(4i+1)[/tex]
[tex]p(-1-2i) = (-1-2i-2)(-1-2i+1-2i)(-1-2i+1+2i)\\=(-3-2i)(-4i)(0)\\=0[/tex]
[tex]p(1+2i) = (1+2i-2)(1+2i+1-2i)(1+2i+1+2i)\\=(-1+2i)(2)(2+4i)[/tex]
[tex]p(2) = (2-2)(2+1-2i)(2+1+2i)\\=(0)(3-2i)(3+2i)\\=0[/tex]
Hence,
The true statements are:
p(-1+2i)=0
p(-1-2i)=0
p(2)=0
Keywords: Polynomials, expressions
Learn more about polynomials at:
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