An algebra problem by shivamani patil

Let $f(x)=x^3+2x^2-3x+5$ and $g(x)=f(x)+15$ :

$g(x)=x^3+2x^2-3x+20=(x+4) \left(x^2-2 x+5\right)$

Completing the square of $x^2-2x+5$ gives:

$x^2-2x+5=0 \Rightarrow (x-1)^2+4=0 \Rightarrow x-1=\pm 2i \Rightarrow x=1 \pm 2i$

It therefore follows that $1+2i$ is a root of $g(x)$ that is $g(1+2i)=0$ :

$f(1+2i)=g(1+2i)-15=0-15=\boxed{-15}$

Given $x^{3}$ +2 $x^{2}$ -3x+5 ----(1)

where x=1+2i ----(2)

x=1+2i is same as x-1=2i

squaring both sides of x-1=2i

i.e $x^{2}$ -2x+1 = 4* $i^{2}$

i.e $x^{2}$ -2x+1 = 4*-1 because $i^{2}$ = -1

i.e $x^{2}$ -2x+1 = -4

i.e $x^{2}$ -2x+5=0------(3)

now divide $x^{3}$ +2 $x^{2}$ -3x+5 by $x^{2}$ -2x+5

we get remainder -15 and quotient x+4

we have dividend = divisor * quotient + remainder

therefore $x^{3}$ +2 $x^{2}$ -3x+5 = [ $x^{2}$ -2x+5 ]*[x+4]+[-15}

as we know that $x^{2}$ -2x+5 = 0 from equation (3)

therefore $x^{3}$ +2 $x^{2}$ -3x+5= 0*[x+4]+[-15]

therefore $x^{3}$ +2 $x^{2}$ -3x+5=-15 when x= 1+2i.