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A polynomial p(x) has 3, 2-i, and √3 as its roots and goes through (0,-90), find p(x)

User BobDroid
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1 Answer

7 votes

Answer:

p(x) = -2x⁵ +14x⁴ -28x³ -12x² +102x -90

Explanation:

Assuming the desired polynomial has real and rational coefficients, each of the complex and irrational roots has a corresponding conjugate. That is, the roots of p(x) must include ...

{3, 2 -i, 2 +i, √3, -√3}

Each of these roots corresponds to a linear factor. Root p corresponds to factor (x -p), for example.

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Factored form

Then the factored p(x) will be ...

p(x) = k(x -3)(x -(2 -i))(x -(2 +i))(x -√3)(x +√3) . . . . for some scale factor k

Simplifying a bit, we have ...

p(x) = k(x -3)((x -2)² +1)(x² -3) = k(x -3)(x² -4x +5)(x² -3)

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Scale factor

At x=0, this evaluates to ...

p(0) = k(0 -3)(0² -4·0 +5)(0² -3) = 45k

We want p(0) = -90, so we require ...

45k = -90

k = -2

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Then the function p(x) factored to integers is ...

p(x) = -2(x -3)(x² -4x +5)(x² -3)

When we multiply this out, we find the standard form equation is ...

p(x) = -2x⁵ +14x⁴ -28x³ -12x² +102x -90

User Woodings
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