Exercise 19 Page 316

14+sqrt(2), 6i

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We are told that the given roots are roots of a polynomial function P(x) that has rational coefficients. 14-sqrt(2) and - 6i To find two additional roots of P(x)=0, let's recall the Conjugate Root Theorem.

Irrational Conjugate Root Theorem
If P(x) is a polynomial with rational coefficients, then any irrational roots of P(x)=0 occur in conjugate pairs.

The above statement tell us that if a + sqrt(b) is an irrational root, then a - sqrt(b) is also a root. Let's use this to find an additional irrational root.

Hypotheses	Conclusion
P(x) has rational coefficients	14+sqrt(2) is also a root of P(x)
14-sqrt(2) is an irrational root of P(x)=0	14+sqrt(2) is also a root of P(x)

To find the remaining root, let's recall the Conjugate Root Theorem.

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

This means that if a + bi is a complex root, then a - bi is also a root. Let's now use this to find the other complex root.

Hypotheses	Conclusion
P(x) has rational, real coefficients	0+6i=6i is also a root of P(x)
- 6i=0-6i is a complex root of P(x)	0+6i=6i is also a root of P(x)

The two additional roots that we can know for certain using the Complex Conjugate Root Theorem are 14+sqrt(2) and 6i.

Subchapter links

Lesson Check p.315

Practice and Problem-Solving Exercises p.316-317

Standardized Test Prep p.317

Mixed Review p.317

Exercise 19 Page 316

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