Question

N=4
I and 2 I are zeros
F(-2)=40

Answer 1

Answer:

So the answer is $y= (x^{4} +5x^{2}+4 )$

Step-by-step explanation:

Given;

$N=4$ , 'I' and '2I' are zeros and

$F(-2)=40$   (equation-1)

Assuming you need real coefficients so you use complex conjugate roots;

$y=a\times(x-i)\times (x+i)\times (x-2i)\times (x+2i)$   Where $y=F(x)$

$y=a\times (x^{2} -i^{2} )\times (x^{2} -4i^{2} )$   (By applying $(a-b)(a+b)=a^{2} -b^{2}$)

$y=a\times(x^{2}+1 )(x^{2}+4 )$    (We know $i^{2} =-1$ )

$y=a\times (x^{4}+4x^{2} +x^{2} +4 )$

$y=a\times (x^{4}+5x^{2} +4 )$  (equation-2)

From equation;

$x=-2$ and $y=40$

Plug 'x' and 'y' value in equation-2,

$40=a\times ((-2)^{4}+5(-2)^{2} +4 )$

$40=a\times (16+20+4)$

$40=a\times 40$

$a=1$

Now equation-2 become;

$y=1\times (x^{4} +5x^{2}+4 )$

∴$y= (x^{4} +5x^{2}+4 )$