document.write( "Question 954247: Good evening! I have two questions today:\r
\n" ); document.write( "\n" ); document.write( "1. Find three consecutive integers such that the product of all three, decreased by the cube of the first, is 33\r
\n" ); document.write( "\n" ); document.write( "2. Find three consectutive integers such that the product of the second and the third, decreased by four times the second, is five more than five times the first \n" ); document.write( "
Algebra.Com's Answer #582787 by addingup(3677)\"\" \"About 
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x*(x+1)*(x+2)-x^3= 33 Reorder terms
\n" ); document.write( "x*(1+x)*(2+x)-x^3= 33 Multiply as follows:
\n" ); document.write( "(1+x)*(2+x) = 2+1x+2x+x^2 So we now have:
\n" ); document.write( "x(2+1x+2x+x^2)- x^3= 33 Add on left:
\n" ); document.write( "x(2+3x+x^2)- x^3= 33 Multiply to eliminate parenthesis
\n" ); document.write( "2x+3x^2+x^3- x^3 = 33 Simplify, subtract on the left
\n" ); document.write( "2x+3x^2 = 33 Now, to get a quadratic equation, subtract 33 on both sides
\n" ); document.write( "2x+3x^2-33 = 0 Factor the equation:
\n" ); document.write( "(x-3)(3x+11) = 0 Solve each side separately:
\n" ); document.write( "x-3= 0; x= 3
\n" ); document.write( "3x+11= 0; 3x= -11; x= -11/3
\n" ); document.write( "Discard the negative. Let's try the other answer which is 3:
\n" ); document.write( "3*(3+1)*3+2)-3^3=
\n" ); document.write( "3*4*5-3^3
\n" ); document.write( "60-3^3= 60-27= 33 We got the right answer: 3\r
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