Question 2: Factorize each of the following expresssions:
\((i)\) \(\;\) \( (x+1)(x+2)(x+3)(x+4)+1\)
Solution:
\(\ \ \ \ ( x+1)( x+2)( x+3)( x+4) +1 \)
\(=[( x+1)( x+4)][( x+3)( x+2)] +1\)
\(=[ x( x+4) +1( x+4)][ x( x+2)\ +\ \)\(3( x+2)] +1\)
\(=[ x^{2} +4x+x+4][ x^{2} +2x+3x+6]+1\)
\(=[ x^{2} +5x+4][ x^{2} +5x+6] +1\)
Let \(\; y=x^2+5x\), we have
\(=[ y+4][ y+6] +1\)
\(=y( y+6) +4( y+6) +1\)
\(=y^{2} +6y+4y+24+1\)
\(=y^{2} +10y+25\)
\(=y^{2} +5y+5y+25\)
\(=y( y+5) +5( y+5)\)
\(=( y+5)( y+5)\)
\(=( y+5)^{2}\)
\(=\left( x^{2} +5x+5\right)^{2}\)
——– Alternate Method ——–
\(\ \ \ \ ( x+1)( x+2)( x+3)( x+4) +1 \)
\(=[( x+1)( x+4)][( x+3)( x+2)] +1\)
\(=[ x( x+4) +1( x+4)][ x( x+2)\ +\ \)\(3( x+2)] +1\)
\(=[ x^{2} +4x+x+4][ x^{2} +2x+3x+6]+1\)
\(=[ x^{2} +5x+4][ x^{2} +5x+6] +1\)
Let \(\; y=x^2+5x\), we have
\(=[ y+4][ y+6] +1\)
\(=y( y+6) +4( y+6) +1\)
\(=y^{2} +6y+4y+24+1\)
\(=y^{2} +10y+25\)
\(=y^{2} +2(y)(5)+(5)^2\)
\(=( y+5)^{2}\)
\(=\left( x^{2} +5x+5\right)^{2}\)