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Question Number 11023 by ABD last updated on 07/Mar/17

(x^3 +6)×P(x)+6x=ax^3 +2ax+b+3  ⇒b=?

$$\left({x}^{\mathrm{3}} +\mathrm{6}\right)×{P}\left({x}\right)+\mathrm{6}{x}={ax}^{\mathrm{3}} +\mathrm{2}{ax}+{b}+\mathrm{3} \\ $$$$\Rightarrow{b}=? \\ $$

Answered by mrW1 last updated on 07/Mar/17

(x^3 +6)×P(x)+6x=ax^3 +2ax+b+3  P(x)x^3 +6×P(x)+6x=ax^3 +2ax+b+3  ⇒P(x)=a  ⇒6=2a ⇒a=3  ⇒6×P(x)=b+3 ⇒b=6×3−3=15

$$\left({x}^{\mathrm{3}} +\mathrm{6}\right)×{P}\left({x}\right)+\mathrm{6}{x}={ax}^{\mathrm{3}} +\mathrm{2}{ax}+{b}+\mathrm{3} \\ $$$${P}\left({x}\right){x}^{\mathrm{3}} +\mathrm{6}×{P}\left({x}\right)+\mathrm{6}{x}={ax}^{\mathrm{3}} +\mathrm{2}{ax}+{b}+\mathrm{3} \\ $$$$\Rightarrow{P}\left({x}\right)={a} \\ $$$$\Rightarrow\mathrm{6}=\mathrm{2}{a}\:\Rightarrow{a}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{6}×{P}\left({x}\right)={b}+\mathrm{3}\:\Rightarrow{b}=\mathrm{6}×\mathrm{3}−\mathrm{3}=\mathrm{15} \\ $$

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