10. a. T or $$ \mathrm{F}: $$ A 3rd degree polynomial can have 3 non-real roots. b. T or $$ \mathrm{F}: \mathrm{f}( ext{x})=( ext{x}-3)^{3}( ext{x}-6) $$ has a zero with multiplicity 3.

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10. a. T or $$ \mathrm{F}: $$ A 3rd degree polynomial can have 3 non-real roots. b. T or $$ \mathrm{F}: \mathrm{f}(	ext{x})=(	ext{x}-3)^{3}(	ext{x}-6) $$ has a zero with multiplicity 3.

UpStudy AI · Accepted Answer

Let's analyze each statement:

**a. True or False: A 3rd degree polynomial can have 3 non-real roots.**

- **Answer:** **False**
  
  **Explanation:** For polynomials with real coefficients, non-real roots must come in complex conjugate pairs. Therefore, a 3rd degree (cubic) polynomial with real coefficients can have at most two non-real roots (which are conjugates) and one real root. It cannot have three non-real roots. If complex coefficients are allowed, then a cubic polynomial could have three non-real roots, but typically, when not specified, polynomials are assumed to have real coefficients.

**b. True or False: $$ f(x) = (x-3)^3(x-6) $$ has a zero with multiplicity 3.**

- **Answer:** **True**
  
  **Explanation:** The polynomial $$ f(x) = (x-3)^3(x-6) $$ has zeros at $$ x = 3 $$ and $$ x = 6 $$. The zero at $$ x = 3 $$ has a multiplicity of 3 because the factor $$ (x-3) $$ is raised to the third power. The zero at $$ x = 6 $$ has a multiplicity of 1.

**Summary:**
- **a. False**
- **b. True**
a. False b. True

a. T or A 3rd degree polynomial can have 3 non-real roots.
b. T or has a zero with multiplicity 3.

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a. T or A 3rd degree polynomial can have 3 non-real roots. b. T or has a zero with multiplicity 3.