^{4}-8x

^{3}+4x

^{2}+4x+39

^{4}-8(3+2i)

^{3}+4(3+2i)

^{2}+4(3+2i)+39

^{2})

^{2}=-1

^{3}

^{2}=(5+12i)

^{2})

^{4}

^{2}

^{4}+5x

^{3}+7x

^{2}-x+41

^{4}+5(-2-√3i)

^{3}+7(-2-√3i)

^{2}-(-2-√3i)+41

^{2})

^{3}

^{2})

^{4}

^{2})

1. If f(x)=x^4 -8x^3+4x^2+4^x+39 and f(3+2i)= a+ib, then a:b equals?

2. If x= -2- root(3)i then value of 2x^4+5x^3+7x^2-x+41 is?

please give explanation in detail, thanks

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Hello Shivam,

f(x)=x^{4} -8x^{3}+4x^{2}+4x+39

(I'm assuming this was supposed to say 4x rather than 4^x based on the context of the problem.)

1. f(3+2i)=(3+2i)^{4}-8(3+2i)^{3}+4(3+2i)^{2}+4(3+2i)+39

I'm going to do each term separately starting with the squared one to keep it simple.

4(3+2i)(3+2i)

=4(9+6i+6i+4i^{2})

Now remember that i^{2}=-1

=4(9+12i-4)

=4(5+12i)

=20+48i

8(3+2i)^{3}

=8(5+12i)(3+2i) Since we already determined that (3+2i)^{2}=(5+12i)

=8(15+10i+36i+24i^{2})

=8(15+46i-24)

=8(-9+46i)

=-72+368i

(3+2i)^{4}

=(5+12i)(5+12i)

=25+60i+60i+144i^{2}

=25+120i-144

=-119+120i

Back to the original:

f(3+2i)=-119+120i-72+368i+20+48i+12+8i+39

=-120+544i

Therefore a=-120 and b=544

2. f(x)=2x^{4}+5x^{3}+7x^{2}-x+41

f(-2-√3i)=2(-2-√3i)^{4}+5(-2-√3i)^{3}+7(-2-√3i)^{2}-(-2-√3i)+41

7(-2-√3i)(-2-√3i)

=7(4-2√3i-2√3i+3i^{2})

=7(4-4√3i-3)

=7(1-4√3i)

=7-28√3i

5(-2-√3i)^{3}

=5(1-4√3i)(-2-√3i)

=5(-2-√3i+8√3i+12i^{2})

=5(-2+7√3i-12)

=5(-14+7√3i)

=-70+35√3i

7(-2-√3i)^{4}

=7(1-4√3i)(1-4√3i)

=7(1-4√3i-4√3i-48i^{2})

=7(1-8√3i+48)

=7(49-8√3i)

=343-56√3i

f(-2-√3i)=343-56√3i-70+35√3i+7-28√3i+2+√3i+41

=323-48√3i

In a long operation like this, there are a lot of opportunities for human errors, so you would be wise to run through it twice.

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