# given the polynomial f=4x^3-12x^2+mx+n find m, n if f(x)=0 for x=i

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Given the polynomial f=4x^3 - 12x^2 + mx + n, we have to find m and n if f(x) = 0 for x= i

Now for x = i

f(i) = 4x^3 - 12x^2 + mx + n

=> 4*i^3 - 12*i^2 + m*i + n = 0

=> -4i + 12 + mi + n =0

=> i( m - 4) + 12 + n = 0

equate the complex and real parts

=> m - 4 = 0

=> m = 4

and 12 + n = 0

=> n = -12

**Therefore m = 4 and n = -12.**

If f(x) = 0 for x =i, that means that x = i is the root of the polynomial.

f(i) = 0

We'll substitute x by i and we'll have:

f(i) = 4*i^3 - 12*i^2 + m*i + n

We'll put i^2 = -1

We'll multiply by i and we'll get:

i^3 = -i

f(i) = 4*(-i) - 12*(-1) + m*i + n

f(i) = -4i + 12 + mi + n

But f(i) = 0

-4i + 12 + mi + n = 0

We'll combine real parts and imaginary parts:

(12 + n) + i(-4 + m) = 0 + 0*i

We'll compare and we'll have:

n + 12 = 0

**n = -12**

m - 4 = 0

**m = 4**

**The polynomial is:**

**f(x) = 4*x^3 - 12*x^2 + 4*x - 12**

Given the polynomial f=4x^3-12x^2+mx+n find m, n if f(x)=0 for x=i.

If f(i) = 0, then i is the root of 4x^3-12x^2+mx+n.

Since complex roots appear in pairs , if i is a root then -i os also a root..

Therefore f(x) the factors, (x+i)(x-i) = x^2+1.

Therefore we can write 4x^3-12x^2+mx+n = (x^2+1)(ax+b).

4x^3-12x^2+mx+n = ax^3+bx^2+ax+b.

We equate the coefficients of like terms:

a = 4, b= =-12,

m = a = 4

n = b = -12.

Therefore m = 4 and n = -12.

Therefore 4x^3-12x^2+mx+n = x^3-12x^2+4x-12.