document.write( "Question 760234: When the polynomials f(x)=(a-1)x^3+ax^2+bx+c where a,b,and c are constants is divided by (x+2) and (x-1),the remainder are -5 and 4 respectively.if (x+1) is a factor of f(x),find the value of a,b,c. Factorise f(x) completely,zeros of f(x)and sketch the graph of f(x). \n" ); document.write( "
Algebra.Com's Answer #462504 by Edwin McCravy(20060)\"\" \"About 
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When the polynomials f(x)=(a-1)x³+ax²+bx+c where a,b,and c are constants is divided by (x+2) and (x-1),the remainder are -5 and 4 respectively.if (x+1) is a factor of f(x),find the value of a,b,c. Factorise f(x) completely,zeros of
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document.write( "If f(x) = (a-1)x³+ax²+bx+c is divided by (x+2), the remainder is -5.\r\n" );
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document.write( "That tells us that f(-2) = -5\r\n" );
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document.write( "f(-2) = (a-1)(-2)³+a(-2)²+b(-2)+c = -5\r\n" );
document.write( "f(-2) = (a-1)(-8)+a(4)-2b+c = -5\r\n" );
document.write( "f(-2) = -8(a-1)+4a-2b+c = -5\r\n" );
document.write( "f(-2) = -8a+8+4a-2b+c = -5\r\n" );
document.write( "f(-2) = -4a+8-2b+c = -5\r\n" );
document.write( "or\r\n" );
document.write( "        -4a-2b+c = -13\r\n" );
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document.write( "f(x)=(a-1)x³+ax²+bx+c where a,b,and c are constants is divided by (x-1), the remainder is 4.\r\n" );
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document.write( "That tells us that f(1) = 4\r\n" );
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document.write( "f(1) = (a-1)(1)³+a(1)²+b(1)+c = 4\r\n" );
document.write( "f(1) = (a-1)+a+b+c = 4\r\n" );
document.write( "f(1) = a-1+a+b+c = 4 \r\n" );
document.write( "f(1) = 2a+b+c-1 = 4\r\n" );
document.write( "or\r\n" );
document.write( "       2a+b+c = 4  \r\n" );
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document.write( "(x+1) is a factor of f(x)\r\n" );
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document.write( "That tells us that f(-1) = 0\r\n" );
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document.write( "f(-1) = (a-1)(-1)³+a(-1)²+b(-1)+c = 0\r\n" );
document.write( "f(-1) = -1(a-1)+a-b+c = 0\r\n" );
document.write( "f(-1) = -a+1+a-b+c = 0\r\n" );
document.write( "f(-1) = 1-b+c = 0\r\n" );
document.write( "or\r\n" );
document.write( "            -b+c = -1\r\n" );
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document.write( " So we have the system of equations:\r\n" );
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document.write( "        -4a-2b+c = -13\r\n" );
document.write( "         2a +b+c =   5 \r\n" );
document.write( "            -b+c =  -1\r\n" );
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document.write( "Solve that and get a=3,b=0, c=-1\r\n" );
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document.write( "So\r\n" );
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document.write( "f(x) = (a-1)x³+ax²+bx+c becomes:\r\n" );
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document.write( "f(x) = (3-1)x³+(3)x²+(0)x+(-1)\r\n" );
document.write( "f(x) = 2x³ + 3x² - 1\r\n" );
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document.write( "Since we are told that (x+1) is a factor, -1 is a zero (or root)\r\n" );
document.write( "and we may divide by synthetic division:\r\n" );
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document.write( "-1|2  3  0 -1\r\n" );
document.write( "  |  =2 -1  1\r\n" );
document.write( "   2  1 -1  0\r\n" );
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document.write( "Thus we have factored (factorised, as you say in the UK), f(x) as\r\n" );
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document.write( "f(x) = (x+1)(2x²+x-1)\r\n" );
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document.write( "and the complete factorization (you spell it factorisation) is\r\n" );
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document.write( "f(x) = (x+1)(2x-1)(x+1)\r\n" );
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document.write( "or since (x+1) is a factor twice,\r\n" );
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document.write( "f(x) = (x+1)²(2x-1)\r\n" );
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document.write( "Edwin
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