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Question Number 206230 by BaliramKumar last updated on 09/Apr/24

Expand x^2 + 2x + 3 respect to x = −2. (a) (x − 2)^2 −2(x + 2) + 3 (b) (x + 2)^2 −2(x + 2) + 3 (c) (x + 2)^2 + 2(x + 2) + 3 (d) (x − 2)^2 −2(x − 2) − 3 is it taylors theorem?

$$\mathrm{Expand}\:\:\:\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{3}\:\:\:{respect}\:{to}\:{x}\:=\:−\mathrm{2}. \\ $$$$\left(\mathrm{a}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{b}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{c}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} +\:\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{d}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:−\:\mathrm{2}\right)\:−\:\mathrm{3} \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{taylors}\:\mathrm{theorem}? \\ $$

Answered by Frix last updated on 10/Apr/24

x^2 +2x+3 x=t−2 t^2 −2t+3 t=x+2 (x+2)^2 −2(x+2)+3

$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3} \\ $$$${x}={t}−\mathrm{2} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{3} \\ $$$${t}={x}+\mathrm{2} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}+\mathrm{2}\right)+\mathrm{3} \\ $$

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