Question

# - IX/40-45400 find value = fa 2+60) (42-45) If 23 _ 5002 PAC+24=18+4): 212) then of P

Solution

${x}^{3}-5{x}^{2}-px+24=(x+4)\xb7q\left(x\right)\phantom{\rule{0ex}{0ex}}Thus,{x}^{3}-5{x}^{2}-px+24isdividedby\phantom{\rule{0ex}{0ex}}x+4andproducesquotientq\left(x\right).\phantom{\rule{0ex}{0ex}}x+4\sqrt{{x}^{3}-5{x}^{2}-px+24}({x}^{2}-9x+6\phantom{\rule{0ex}{0ex}}{x}^{3}+4{x}^{2}\phantom{\rule{0ex}{0ex}}\overline{)-9{x}^{2}-px}\phantom{\rule{0ex}{0ex}}-9{x}^{2}\pm 36x\phantom{\rule{0ex}{0ex}}\overline{)\left(36-p\right)x+24}\phantom{\rule{0ex}{0ex}}6x+24\phantom{\rule{0ex}{0ex}}\overline{)0}\phantom{\rule{0ex}{0ex}}Thus36-p=6\phantom{\rule{0ex}{0ex}}\mathit{p}\mathbf{=}\mathbf{36}\mathbf{-}\mathbf{6}\mathbf{=}\mathbf{30}$