document.write( "Question 75456: Find the linear factorization from the given zeros
\n" ); document.write( "Zeros: -1, multiplicity 1; 2, multiplicity 3; 4, multiplicity 2 \n" ); document.write( "

Algebra.Com's Answer #54329 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
If we are given a zero of any value (say x=a) then the factor will be
\n" ); document.write( " $\"x-a=0\"$ . The reason why is if we solved for x we get
\n" ); document.write( " $\"x=a\"$ . If we have 2 zeros of x=a and x=b, then the factors are
\n" ); document.write( " $\"%28x-a%29%28x-b%29=0\"$ See the pattern?
\n" ); document.write( "Simply if we are given zeros, we can easily find the polynomial. So the factors of the polynomial are\r
\n" ); document.write( "\n" ); document.write( " $\"%28x%2B1%29%28x-2%29%28x-4%29\"$
\n" ); document.write( "Since the 2nd factor has a multiplicity of 3, we cube the 2nd term. Also, since the 3rd term has a multiplicity of 2, we simply square the term
\n" ); document.write( " $\"%28x%2B1%29%28x-2%29%5E3%28x-4%29%5E2\"$
\n" ); document.write( "And there's the linear factorization. If you multiplied all the terms out, you would get a polynomial with the same zeros as given in the problem. \n" ); document.write( "

\n" );