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2 years ago

PLEASE HELP ME!!!!! (Urgent)

Mathematics

2 answers:

defon2 years ago

8 0

Answer:

$f(x)=x^5-6x^4+14x^3-24x^2+40x$

Step-by-step explanation:

<u>Given information</u>:

Polynomial function with real coefficients.
Zeros: 0, 2i and (3+i).

For any complex number $z=a+bi$ , the complex conjugate of the number is defined as $z^*=a-bi$ .

If f(z) is a polynomial with real coefficients, and z₁ is a root of f(z)=0, then its complex conjugate z₁* is also a root of f(z)=0.

Therefore, if f(x) is a polynomial with real coefficients, and 2i is a root of f(x)=0, then its complex conjugate -2i is also a root of f(x)=0.

Similarly, if (3+i) is a root of f(x)=0, then its complex conjugate (3-i) is also a root of f(x)=0.

Therefore, the polynomial in factored form is:

$f(x)=ax(x-2i)(x-(-2i))(x-(3+i))(x-(3-i))$

$f(x)=ax(x-2i)(x+2i)(x-3-i)(x-3+i)$

As we have not been given a leading coefficient, assume a = 1:

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

Expand the polynomial:

$f(x)=x(x^2+2ix-2ix-4i^2)(x^2-3x+xi-3x+9-3i-xi+3i-i^2)$

$f(x)=x(x^2-4i^2)(x^2-6x+9-i^2)$

$f(x)=x(x^2-4(-1))(x^2-6x+9-(-1))$

$f(x)=x(x^2+4)(x^2-6x+10)$

$f(x)=(x^3+4x)(x^2-6x+10)$

$f(x)=x^5-6x^4+10x^3+4x^3-24x^2+40x$

$f(x)=x^5-6x^4+14x^3-24x^2+40x$

Vsevolod [243]2 years ago

6 0

The polynomial function f(x) = x³ - 6x² - 8x

<h3>How to write the polynomial function?</h3>

To find a polynomial function f(x) of least degree having only real coefficients with zeros of 0, 2i, and 3+i, its factors are x, x - 2i and x - (3 + i)

So f(x) is the product of the factors

f(x) = x(x - 2i)[x - (3 + i)]

= (x² - 2ix)[x - (3 + i)]

= x³ - x²(3 + i) - 2ix² - 2ix × (-[3 + i])]

= x³ - 3x² - x²i - 2ix² - 2ix × (-[3 + i])]

= x³ - 3x² - x²i - 2ix² - 6ix + 2i²x

= x³ - 3x² - x²i - 2ix² - 6ix - 2x

= x³ - 3x² - x²i × -i - 2ix² × -i - 6ix × -i - 2x

= x³ - 3x² - x²i × -i - 2(-i²)x² - 6(-i²)x - 2x

= x³ - 3x² - x² - 2x² - 6x - 2x

= x³ - 3x² - 3x² - 8x

= x³ - 6x² - 8x