document.write( "Question 628003: Find a polynomial f(x) with real coefficients having the given degree and zeros.
\n" ); document.write( "Degree 5; zeros: 6;-i;-7+i\r
\n" ); document.write( "\n" ); document.write( "f(x)=a?
\n" ); document.write( "Do not write the polynomial in factored form. Multiply it out. \n" ); document.write( "

Algebra.Com's Answer #395294 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "If

is a zero of a polynomial, then

is a factor of the polynomial. Complex zeros of polynomials with real coefficients always appear in conjugate pairs. You didn't specify that your function has real coefficients, but unless that is true it is impossible to solve the problem as posed. Hence, I make the assumption of real coefficients.\r
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is a real zero, so

is a factor of the polynomial. The conjugate of a general complex number

is formed by changing the sign on the second term thus:

is a complex zero. To form the conjugate of

, rewrite it as

, then it should be clear that the conjugate is

which can be written as simply

. Likewise your other zero,

, is a complex number, hence the fifth and last zero of your 5th degree polynomial is

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\n" ); document.write( "\n" ); document.write( "Now just form the factors:\r
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\r
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\n" ); document.write( "\n" ); document.write( "Now the only work left for you to do is to multiply out the five binomials. Note that the product of two conjugate binomials is the difference of two squares, but because

, the product of two complex conjugates becomes the sum of two squares.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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$\"The$

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