document.write( "Question 579150: find zeros and multiplicities
\n" ); document.write( "a) f(x)= x^3-12x^2-55x+150 \n" ); document.write( "

Algebra.Com's Answer #370916 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"f%28x%29=+x%5E3-12x%5E2-55x%2B150\"$ is a polynomial
\n" ); document.write( "We know that if it has a rational zero, it will be a \"fraction\" whose denominator is a factor of the leading coefficient and whose numerator is a factor of the constant (or independent term, whatever you call it), with a plus or minus sign in front.
\n" ); document.write( "The leading coefficient (the coefficient of the term of highest degree) is that invisible 1 in front of $\"x%5E3\"$ . So the denominator is a factor of 1, which could only be 1. Any rational zero will be an integer. It has to be a factor of 150. There are 12 positive factors of 150: 1, 2, 3, 5, 6, 10, 15, 25, 30,7 5, and 150. The same numbers with a minus sign in front are also possibilities. I tried 1, and it was not a zero, but then I tried 2, and 2 was a zero.
\n" ); document.write( "Since 2 is a zero of $\"f%28x%29\"$ , $\"%28x-2%29\"$ is a factor of $\"f%28x%29\"$ . $\"f%28x%29\"$ can be evenly divided by $\"%28x-2%29\"$ .
\n" ); document.write( "I did the division and the result was $\"x%5E2-10-75\"$ . So
\n" ); document.write( " $\"f%28x%29=+x%5E3-12x%5E2-55x%2B150=%28x-2%29%28x%5E2-10-75%29\"$
\n" ); document.write( "Factoring $\"x%5E2-10-75\"$ , I found that $\"x%5E2-10-75=%28x-15%29%28x%2B10%29\"$
\n" ); document.write( "So $\"f%28x%29=+%28x-2%29%28x-15%29%28x%2B10%29\"$ and its zeros are x=2, x=15, and x=-10. \n" ); document.write( "

\n" );