How do you write a polynomial in standard form given zeros -1 (multiplicity 2), -2 - i (multiplicity 1)?

1 Answer
Sep 29, 2016

Answer:

#f(x)=x^4+6x^3+14x^2+14x+5#

Explanation:

Zeros:
#-1# multiplicity 2
#(-2-i)# multiplicity 1

If the zero is #-1#, the factor is #(x-(-1))=(x+1)#.
A multiplicity of #color(red)2# implies there are #color(red)2# factors #(x+1)# or #(x+1)(x+1)= (x+1)^color(red)2 =(x^2+2x+1)#

For the complex zero #(-2-i)#, there will also be a zero at the complex conjugate#(-2+i)#

The factors are #(x-(-2-i))# and #(x-(-2+i))#
or
#(x+2+i)(x+2-i)#
#(x^2+2x-ix+2x+4-2i+ix+2i-i^2)#
#(x^2+4x+4-(-1))#
#(x^2+4x+5)#

Multiply all factors to find the polynomial in standard form
#f(x)=(x^2+2x+1)(x^2+4x+5)=#

#x^4+4x^3+5x^2#
#color(white)(aaaa)2x^3+8x^2+10x#
#color(white)(aaaaaaaaaa)x^2+4x+5=#

#f(x)=x^4+6x^3+14x^2+14x+5#