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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2Let \( \mathrm{{\mathbf{{P}}}} \)be the space of all polynomials.Now we may prove why \( \mathrm{{\mathbf{{P}}}} \) is an infinite denominational vector space. Suppose the dimension of \( \mathrm{{\mathbf{{P}}}} \) is finite say n. The linear independent contains the set of more than n vectors.Consider the set \( \mathrm{{S}={\left\lbrace{1},{x},{x}^{{2}},{x}^{{3}},---,{x}^{{n}}\right\rbrace}} \)Let \( \mathrm{{a}_{{0.1}}+{a}_{{1}}.{x}+{a}_{{2}}.{x}^{{2}}+\ldots+{a}_{{n}}.{x}^{{n}}={0}} \)Then\( \mathrm{{x}={0}} \) implies \( \mathrm{{a}_{{0}}={0}} \)Therefore \( \mathrm{{a}_{{1}}.{x}+{a}_{{2}}.{x}^{{2}}+\ldots+{a}_{{n}}.{x}^{{n}}={0}} \)Then \( \mathrm{{a}_{{1}}+{a}_{{2}}.{x}+{a}_{{3}}.{x}^{{2}}+\ldots+{a}_{{n}}.{x}^{{{n}-{1}} ... See the full answer