document.write( "Question 1173289: If (2+2) and (x-3) are factors of the polynomial f(x) =x^2+ax^2-7x+b where a and b are constant determine the values of a and b and hence factorise f(x) completely
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Algebra.Com's Answer #798505 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "I think you made two errors. Tou wrote this:\r\n" );
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\n" ); document.write( "If (2+2) and (x-3) are factors of the polynomial f(x) =x^2+ax^2-7x+b where a
\n" ); document.write( "and b are constant determine the values of a and b and hence factorise f(x)
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document.write( "I think you meant this:

\n" ); document.write( "If (x+2) and (x-3) are factors of the polynomial f(x) =x^3+ax^2-7x+b where a
\n" ); document.write( "and b are constant determine the values of a and b and hence factorise f(x)
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document.write( "Can you find your two errors?\r\n" );
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document.write( "But instead of doing that one for you I will do this one, which is exactly like yours step by step.

\n" ); document.write( "If (x+7) and (x-4) are factors of the polynomial f(x) =x^3+ax^2-13x+b where a
\n" ); document.write( "and b are constant determine the values of a and b and hence factorise f(x)
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document.write( "Since they are both factors of f(x), their product is also a factor of f(x).\r\n" );
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document.write( "We multiply them together: (x+7)(x-4) = x²-4x+7x-28 = x²+3x-28.\r\n" );
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document.write( "Since that product is a factor we divide f(x) by it using long division. You\r\n" );
document.write( "will have to spread the long division out like this since the coefficients\r\n" );
document.write( "involve a and b:\r\n" );
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document.write( "                                         x +     (a-3)\r\n" );
document.write( "      x²+3x-28)x³+         ax²        -13x +         b\r\n" );
document.write( "               x³+         3x²        -28x\r\n" );
document.write( "                       (a-3)x²         15x           b\r\n" );
document.write( "                       (a-3)x²     (3a-9)x -   28(a-3)\r\n" );
document.write( "                              [15-(3a-9)]x + b+28(a-3)\r\n" );
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document.write( "The quotient is x+(a-3).\r\n" );
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document.write( "The remainder must be identically 0 for all values of x, so:\r\n" );
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document.write( "\"15-%283a-9%29=0\"  and \"b%2B28%28a-3%29=0\"\r\n" );
document.write( "\"15-3a%2B9=0\"\r\n" );
document.write( "\"24-3a=0\"\r\n" );
document.write( "\"24=3a\"\r\n" );
document.write( "\"8=a\"\r\n" );
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document.write( "Substitute a=8 in \"b%2B28%28a-3%29=0\"\r\n" );
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document.write( "\"b%2B28%288-3%29=0\"\r\n" );
document.write( "\"b%2B28%285%29=0\"\r\n" );
document.write( "\"b%2B140=0\"\r\n" );
document.write( "\"b=-140\"\r\n" );
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document.write( "Therefore the quotient is x + (a-3) = x + (8-3) = x+5\r\n" );
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document.write( "So the complete factorisation is (x+7)(x-4)(x+5). \r\n" );
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document.write( "Now do your problem the same way.\r\n" );
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document.write( "Edwin
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