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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

Chapter Review

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Exercise 54 Page 350

If P(x) is a polynomial with real coefficients, then the complex roots of P(x)=0 occur in conjugate pairs.

P(x)=x^2-12x+37

Practice makes perfect

We want to write a polynomial function with rational coefficients so that P(x)=0 has the given root. 6-i To do so, recall the Conjugate Root Theorem for complex roots.

Conjugate Root Theorem
If P(x) is a polynomial with real coefficients, then the complex roots of P(x)=0 occur in conjugate pairs.

This theorem states that if a + bi is a complex root, then a- bi is also a root.

Root	Factor
6-i	x-(6-i)
6+i	x-(6+i)
Polynomial	P(x)= (x-(6-i)) (x-(6+i))

By applying the conjugate root theorem, we find the following roots for the polynomial. (x-(6-i)) * (x-(6+i))

Let's simplify the polynomial by applying the Distributive Property.

(x-(6-i))(x-(6+i))

Distribute - 1

(x-6+i)(x-6-i)

Distribute (x-6-i)

x(x-6-i)-6(x-6-i)+i(x-6-i)

▼

Distribute x & - 6 & i

Distribute x

x^2-6x-ix-6(x-6-i)+i(x-6-i)

Distribute - 6

x^2-6x-ix-6x+36+6i+i(x-6-i)

Distribute i

x^2-6x-ix-6x+36+6i+ix-6i-i^2

i^2=- 1

x^2-6x-ix-6x+36+6i+ix-6i-(-1)

a-(- b)=a+b

x^2-6x-ix-6x+36+6i+ix-6i+1

Add and subtract terms

x^2-12x+37

Subchapter links

Exercises p.347-352