# Find the polynomial having roots at 0, 2i and 3 - i, and passing through (1, –5).

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### 2 Answers

We have to find the polynomial which has roots at -2i and 3 + i and which passes through (1, -5). We know that if a polynomial has the complex root a + ib, it also has the root a – ib. Therefore, the polynomial here has the roots 0, -2i, + 2i, 3 + i and 3 – i. So it is a polynomial of degree 5.

The polynomial is equal to a*(x – 0) (x – 2i) (x + 2i) (x – 3 – i) (x – 3 +i)

=> a*x(x^2 + 4) [(x – 3) ^2 +1)

=> a*x(x^2 +4) (x^2 + 9 – 6x + 1)

=> a*x(x^2 + 4) (x^2 – 6x + 10)

=> a*x(x^4 – 6x^3 + 10 x^2 + 4x^2 – 24x + 40)

=> a*(x^5 – 6x^4 + 14x^3 – 24x^2 + 40x)

Also, the polynomial passes through (1, -5)

=> -5 = a*(1^5 – 6*1^4 + 14*1^3 – 24*1^2 + 40*1)

=> -5 = a*(1 – 6 + 14 – 24 + 40)

=> -5 = a*25

=> a = -1/5

**Therefore the polynomial is (-1/5)*(x^4 – 6x^3 + 14x^2 – 24x + 40)**

A polynomial which has the complex roots 2i and 3-i, must have also the roots which are the conjugates of 2i and 3-i.

Therefore the give polynomial has the a ral root 0, and complex roots 2i and -2i, 3-i and 3+i.

If a, b, c, are the roots of a polynomial y(x), then y(x) = k(x-a)(x-b)(x-c).

Therefore y(x) = k(x-0)(x-2i)(9x-2i){x+3-i)(x-(3+i)) is the polynomial with roots 0, 2i, -2i, 3-i and 3+i.

y(x) = kx(x^2+4)((x-3)^2+1).

y(x) = kx(x^2+4)(x^2-6x+10). This polynomial should pass through (1-5).

Therefore substituting (x ,y) = (1, -5) in y(x) = kx(x^2+2)(x^2-6x+10), we get:

-5 = k(1)(1+4)((1-6+10).

-5 = k*25.

Therefore k = -5/25= -1/5.

Therefore the required polynomial which has roots, 0, 2i, and 3-i and passes through the point (1,-5) is given by:

y(x) = (-1/5)x{x^2+4)(x^2-6x+10).