document.write( "Question 71858: Here is another one that is not explained in the 8 books that I have. At least not good enough whereas I can make sense of it.\r
\n" ); document.write( "\n" ); document.write( "the degree three polynomial f(x) with real coefficients and leading coefficient 1, has 4 and 3 + i among its roots. Express f(x) as a product of linear and quadratic polynomials with real coefficients.\r
\n" ); document.write( "\n" ); document.write( "Possible answers:
\n" ); document.write( "f(x) =(x+4)(x^2 + 6x +10)
\n" ); document.write( "f(x) =(x-4)(x^2 -6x -9)
\n" ); document.write( "f(x) =(x-4)(x^2 -6x +10)
\n" ); document.write( "f(x) = (x-4)(x^2 -6x +9)\r
\n" ); document.write( "\n" ); document.write( "Thanks to anyone who can possibly explain this to me. \n" ); document.write( "
Algebra.Com's Answer #51372 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
To find the roots (where the graph crosses the x-axis) you must set f(x) (or y) equal to zero. This works because all of the zeros will occur when y=0. So lets say we have a quadratic that we can factor we can easily pull the roots out. For instance lets say we want to find the zeros of
\n" ); document.write( "\"x%5E2%2B6x%2B5=0\"We can factor this to
\n" ); document.write( "\"%28x%2B5%29%28x%2B1%29=0\"If we divide both sides by (x+5) we get
\n" ); document.write( "\"x%2B1=0\"Now solve for x
\n" ); document.write( "\"x=-1\"There's one zero, you can verify this if you graph \"x%5E2%2B6x%2B5=0\"
\n" ); document.write( "Now go back to the original product of factors
\n" ); document.write( "\"%28x%2B5%29%28x%2B1%29=0\"Divide both sides by (x+1)
\n" ); document.write( "\"x%2B5=0\"Solve for x
\n" ); document.write( "\"x=-5\"Theres the other zero
\n" ); document.write( "So the zeros for this example are (-5,0) and (-1,0)
\n" ); document.write( "In a sense, if we have ab=0 where a and b are factors, then a or b can equal zero (or both could be zero). If either are zero then the entire equation equals zero. This is why you have to factor a sum (which is what a polynomial is) into a product of linear factors.
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\n" ); document.write( "Now we want to go backwards. We want to take the zeros and find the product. If we know that 4 is a root, then we can solve for one of the factors. In this case (x-4) is the root since if we plugged in 4 we'd get 0. So this eliminates half of the choices. To find the other root, we're going to have to plug in the other factor into the quadratic equation.
\n" ); document.write( "\"%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a\"
\n" ); document.write( "\"%286%2B-sqrt%286%5E2-4%281%29%2810%29%29%29%2F2\"Lets start with the factor (x^2-6x+10)
\n" ); document.write( "\"%286%2B-sqrt%2836-40%29%29%2F2\"
\n" ); document.write( "\"%286%2B-sqrt%28-4%29%29%2F2\"Now since the radicand (the stuff under the square root) is negative, you won't get a real number. However to solve this equation, complex numbers are introduced to find the roots.
\n" ); document.write( "\"%286%2B-sqrt%284%29sqrt%28-1%29%29%2F2\"Factor out a -1 out of the square root. The square root of -1 is not possible, but in order to represent it, the letter i (for imaginary) will stand in its place
\n" ); document.write( "\"%286%2B2%2Ai%29%2F2\" and \"%286-2%2Ai%29%2F2\" (there are two answers since its both positive and negative)
\n" ); document.write( "\"3%2Bi\" Simplify the positive answer
\n" ); document.write( "Since 3+i is a root, this shows that f(x) =(x-4)(x^2 -6x +10) is your answer. I hope that helps. I'm assuming you're mainly having trouble with the complex root, so feel free to ask more questions.\r
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