Question Number 172704 by mathlove last updated on 30/Jun/22
$${if}\:\left({x}^{\mathrm{2}} +\mathrm{11}{x}\right)+\frac{\mathrm{5}}{\left({x}^{\mathrm{2}} −{x}\right)}=−\mathrm{30} \\ $$$${then}\:{faind}:\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)=? \\ $$
Answered by mahdipoor last updated on 30/Jun/22
![⇒x≠0,1 & A=(x+2)(x+3)=x^2 +5x+6 ; ⇒(x^2 −x)(x^2 +11x)+5=−30(x^2 −x) ⇒(x^2 −x)(x^2 +11x+30)=−5 ⇒[x(x+5)].[(x−1)(x+6)]=−5= (x^2 +5x)(x^2 +5x−6)=(A−6)(A−12) ⇒ A=11 or 7](https://www.tinkutara.com/question/Q172710.png)
$$\Rightarrow{x}\neq\mathrm{0},\mathrm{1}\:\&\:{A}=\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)={x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6} \\ $$$$; \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −{x}\right)\left({x}^{\mathrm{2}} +\mathrm{11}{x}\right)+\mathrm{5}=−\mathrm{30}\left({x}^{\mathrm{2}} −{x}\right) \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −{x}\right)\left({x}^{\mathrm{2}} +\mathrm{11}{x}+\mathrm{30}\right)=−\mathrm{5} \\ $$$$\Rightarrow\left[{x}\left({x}+\mathrm{5}\right)\right].\left[\left({x}−\mathrm{1}\right)\left({x}+\mathrm{6}\right)\right]=−\mathrm{5}= \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{5}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{6}\right)=\left({A}−\mathrm{6}\right)\left({A}−\mathrm{12}\right) \\ $$$$\Rightarrow\:{A}=\mathrm{11}\:{or}\:\mathrm{7} \\ $$
Commented by mathlove last updated on 30/Jun/22
$${thanks} \\ $$