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r is a zero of polynomial f(x) with multiplicity 2 and of polynomial g(x) with multiplicity 3. By the factor theorem, f(x) and g(x) can be written as f(x) = k(x)*(x-r)^2 and g(x) = p(x)*(x-r)^3. Squaring both sides gives f^2(x) = k^2(x)*(x-r)^4 and g^2(x) = p^2(x)*(x-r)^6. Therefore, h(x) = f^2(x) + g^2(x) = (x-r)^4(k

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- 1. b) If r has multiplicity 2 as a zero of polynomial f(x) and multiplic- ity 3 as a zero of polynomial g(x). Explain why it is also a zero of h(x) = f2(x) + g2(x). Can we determine it's multiplicity? If so what is the multiplicity? If not, why not? the response should be related to Fundamental theorem of algebra 1 or 2, or remainder theorem factore theorem or rational root theorem- gotta be related to something Solution It is realted to factor theorem If r is a zero for f(x) with multiplicity m then (x-r)m is a factor of f(x) Hence f(x) = k(x)*(x-r)2 g(x) = p(x)*(x-r)3 => f2(x) = k2(x)*(x-r)4 g2(x) = p2(x)*(x-r)6 =>h(x) = f2(x) + g2(x) = (x-r)4 ( k2(x)+ p2(x)*(x-r)2) Hence r is also a zero of h(x) with multiplicity 4 as (x-r)4 is a factor of h(x)