document.write( "Question 166513: For x²(a-b)+a²(b-x)+b²(x-a) show that (x-a) is a linear factor.
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document.write( "PLEASE HELP ME!!!!!! \n" );
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Algebra.Com's Answer #122803 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! For x²(a-b)+ a²(b-x)+ b²(x-a) show that (x-a) is a linear factor. \n" ); document.write( ": \n" ); document.write( "Multiply to eliminate the brackets, we can leave the last term alone \n" ); document.write( "x^2a - x^2b + a^2b - xa^2 + b^2(x-a) \n" ); document.write( ": \n" ); document.write( "Regroup: \n" ); document.write( "x^2a - xa^2 + a^2b - x^2b + b^2(x-a) \n" ); document.write( ": \n" ); document.write( "Factor out xa and b \n" ); document.write( "xa(x-a) + b(a^2 - x^2) + b^2(x-a) \n" ); document.write( ": \n" ); document.write( "Change the sign of b to -, to change signs inside the brackets \n" ); document.write( "xa(x-a) - b(x^2 - b^2) + b^2(x-a) \n" ); document.write( ": \n" ); document.write( "Note that x^2 - b^2 is the \"difference of squares and can be factored \n" ); document.write( "xa(x-a) - b(x-a)(x+a) + b^2(x-a) \n" ); document.write( ": \n" ); document.write( "Note that each term contains the factor (x-a) now \n" ); document.write( " \n" ); document.write( " |