document.write( "Question 30145: x^3+3x^2-25x-75=0\r
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Algebra.Com's Answer #16886 by sdmmadam@yahoo.com(530)\"\" \"About 
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x^3+3x^2-25x-75=0 ----(1)
\n" ); document.write( "(x^3+3x^2)-(25x+75)=0 by additive associativity
\n" ); document.write( "x^2(x+3)-25(x+3) = 0
\n" ); document.write( "x^2p - 25p = 0 where p = (x+3)
\n" ); document.write( "p(x^2-25) =0
\n" ); document.write( "(x+3)(x^2-25) = 0
\n" ); document.write( "(x+3) = 0 implies x = -3
\n" ); document.write( "(x^2-25) = 0 gives x^2 = 25
\n" ); document.write( "Taking sqrt we get x =+5 or x =-5
\n" ); document.write( "Therefore Answer: x = -5,-3 and +5
\n" ); document.write( "Verification: x =-5 in x^3+3x^2-25x-75=0 ----(1)
\n" ); document.write( "LHS = (-5)^3+3X(-5)^2-25X(-5)-75
\n" ); document.write( "=-125+75+125-75
\n" ); document.write( "=(125-125)+(75-75)
\n" ); document.write( "0+0
\n" ); document.write( "=0 =RHS
\n" ); document.write( "x =5 in x^3+3x^2-25x-75=0 ----(1)
\n" ); document.write( "LHS = (5)^3+3X(5)^2-25X(5)-75
\n" ); document.write( "=125+75-125-75
\n" ); document.write( "=(125+75)-(125+75)
\n" ); document.write( "=0 =RHS
\n" ); document.write( "x =-3 in x^3+3x^2-25x-75=0 ----(1)
\n" ); document.write( "LHS = (-3)^3+3X(-3)^2-25X(-3)-75
\n" ); document.write( "=-27+27+75-75
\n" ); document.write( "=0 =RHS
\n" ); document.write( "So all the values hold\r
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