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Find a fourth degree polynomial equation with integral coefficients having the given roots: 2 + I, 1, and- 1/2

User Dj Bazzie Wazzie
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1 Answer

14 votes
14 votes

we know the roots

x=2+i

x=1

x=-1/2

By the conjugate theorem

The fourth root is x=2-i

Write the polynomial in factored form

The factors are

x=2+i --------> [x-(2+i)]

x=2-i---------> [x-(2-i)]

x=1 --------> (x-1)

x=-1/2 --------->( x+1/2)

so

f(x)=a[x-(2+i)]* [x-(2-i)]*(x-1)*(x+1/2)

where

a is the leading coefficient

Multiply [x-(2+i)]* [x-(2-i)]

[x-(2+i)]* [x-(2-i)]=x^2-x(2-i)-x(2+i)+4-i^2=x^2-2x+xi-2x-xi+4+1

so

x^2-4x+5

substitute

f(x)=a(x^2-4x+5)(x-1)(x+1/2)

Multiply (x-1)(x+1/2)

(x-1)(x+1/2)=x^2+0.5x-x-0.5=x^2-0.5x-0.5

substitute

f(x)=a(x^2-4x+5)(x^2-0.5x-0.5)

Multiply

f(x)=a[x^4-0.5x^3-0.5x^2-4x^3+2x^2+2x+5x^2-2.5x-2.5]

f(x)=a[x^4-4.5x^3+6.5x^2-0.5x-2.5]

For a=1

The answer is

f(x)=x^4-4.5x^3+6.5x^2-0.5x-2.5

User Sagar Pise
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2.8k points