document.write( "Question 384334: Factor x^3+2x^2-5x-6 given that x+3 ia a factor. \n" ); document.write( "

Algebra.Com's Answer #272100 by jsmallt9(3758) $\"\"$ $\"About$

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$\"x%5E3%2B2x%5E2-5x-6\"$
\n" ); document.write( "If x+3 is a factor then it will divide evenly into $\"x%5E3%2B2x%5E2-5x-6\"$ . So we can find the other factor by dividing $\"x%5E3%2B2x%5E2-5x-6\"$ by x+3. We can use long division or synthetic division. I find synthetic division easier. But synthetic division works for divisors in the form x-c where c is some number. So to use synthetic division with x+3 we have to look at it as a subtraction: x - (-3):
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document.write( "-3 |  1   2   -5   -6\r\n" );
document.write( "----     -3    3    6\r\n" );
document.write( "      ---------------\r\n" );
document.write( "      1  -1   -2    0\r\n" );
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\n" ); document.write( "The zero in the lower right corner is the remainder and a zero remainder means it (x - (-3)) divided evenly (as we were told it would). The rest of the bottom line tells us the quotient (which is the other factor). The 1 -1 -2 translates into $\"x%5E2-x-2\"$

\n" ); document.write( " $\"x%5E2-x-2\"$ factors fairly easily into (x-2)(x+1). So
\n" ); document.write( " $\"x%5E3%2B2x%5E2-5x-6+=+%28x%2B3%29%28x-2%29%28x%2B1%29\"$ \n" ); document.write( "

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