Use the linear factorization theorem to construct a polynomial function with the given condition.

n=3 with real coefficients, zeros = -4, 2+3i.

n=3 with real coefficients, zeros = -4, 2+3i.

opatovaL
2020-11-08
Answered

n=3 with real coefficients, zeros = -4, 2+3i.

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Nola Robson

Answered 2020-11-09
Author has **94** answers

Step 1

Given:

f(x) is a polynomial with real coefficients of degree 3

Zeroes of f(x) are -4,2 + 3i

Since f(x) is a polynomial with real coefficients the complex roots exist as complex cojugates

So, 2-3i is a zero of f(x) as 2+3i is a zero.

Step 2

Hence, f(x)=(x-4)(x-(2+3i))(x-(2-3i))

f(x)=(x-(-4))((x-2)-3i)((x-2)+3i)

$f\left(x\right)=(x+4)({(x-2)}^{2}+9)$

$f\left(x\right)=(x+4)({x}^{2}-4x+13)$

$f\left(x\right)={x}^{3}-3x+52$

Hence, the polynomial is${x}^{3}-3x+52$

Given:

f(x) is a polynomial with real coefficients of degree 3

Zeroes of f(x) are -4,2 + 3i

Since f(x) is a polynomial with real coefficients the complex roots exist as complex cojugates

So, 2-3i is a zero of f(x) as 2+3i is a zero.

Step 2

Hence, f(x)=(x-4)(x-(2+3i))(x-(2-3i))

f(x)=(x-(-4))((x-2)-3i)((x-2)+3i)

Hence, the polynomial is

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