Question Number 11024 by ABD last updated on 07/Mar/17
![P(x+1)×P(x−1)=4x^2 +8x+a−5 a=?](https://www.tinkutara.com/question/Q11024.png)
$${P}\left({x}+\mathrm{1}\right)×{P}\left({x}−\mathrm{1}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{x}+{a}−\mathrm{5} \\ $$$${a}=? \\ $$
Answered by ajfour last updated on 07/Mar/17
![let P(x) = bx+c Then P(x+1)×P(x−1) =(bx+b+c)(bx−b+c) =(bx+c)^2 −b^2 = 4(x+1)^2 +a−9 so, b=c=±2. and a−9=−b^2 thus a−9 = −4 a=5 .](https://www.tinkutara.com/question/Q11027.png)
$${let}\:{P}\left({x}\right)\:=\:{bx}+{c} \\ $$$${Then}\:{P}\left({x}+\mathrm{1}\right)×{P}\left({x}−\mathrm{1}\right) \\ $$$$=\left({bx}+{b}+{c}\right)\left({bx}−{b}+{c}\right) \\ $$$$=\left({bx}+{c}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} \:\:\:=\:\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{a}−\mathrm{9} \\ $$$${so},\:{b}={c}=\pm\mathrm{2}.\:{and}\:{a}−\mathrm{9}=−{b}^{\mathrm{2}} \\ $$$${thus}\:\:{a}−\mathrm{9}\:=\:−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{a}=\mathrm{5}\:\:. \\ $$