x ( m + 3 ) 2 = 4 x + m + 1 {\displaystyle x(m+3)^{2}=4x+m+1}
x ( m + 3 ) 2 = 4 x + m + 1 x ( m 2 + 6 m + 9 ) − 4 x = m + 1 x ( m 2 + 6 m + 9 − 4 ) = m + 1 x ( m 2 + 6 m + 5 ) = m + 1 x ( m + 5 ) ( m + 1 ) = m + 1 a = ( m + 5 ) ( m + 1 ) b = m + 1 {\displaystyle {\begin{aligned}x(m+3)^{2}=4x+m+1\\x(m^{2}+6m+9)-4x=m+1x(m^{2}+6m+9-4)=m+1\\x(m^{2}+6m+5)=m+1\\x(m+5)(m+1)=m+1\\a=(m+5)(m+1)\ \ b=m+1\\\end{aligned}}}
A . a = ( m + 5 ) ( m + 1 ) ≠ 0 m ≠ − 5 , − 1 B . a = ( m + 5 ) ( m + 1 ) = 0 a n d b = m + 1 ≠ 0 B a ) m = − 5 , − 1 a n d B b ) m ≠ − 1 B b ) m = − 5 C . a = ( m + 5 ) ( m + 1 ) = 0 a n d b = m + 1 = 0 C a ) m = − 5 , − 1 a n d B b ) m = − 1 C b ) m = − 1 {\displaystyle {\begin{aligned}A.a=(m+5)(m+1)\neq 0\\m\neq -5,-1\end{aligned}}\ \ \ \ \ \ {\begin{aligned}B.a=(m+5)(m+1)=0\ and\ b=m+1\neq 0\\Ba)m=-5,-1\ \ \ and\ \ Bb)\ m\neq -1\\Bb)m=-5\\\end{aligned}}\ \ \ \ \ \ {\begin{aligned}C.a=(m+5)(m+1)=0\ and\ b=m+1=0\\Ca)m=-5,-1\ \ \ and\ \ Bb)\ m=-1\\Cb)m=-1\\\end{aligned}}}
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