Tim S.

asked • 07/22/17

I've been at it all day and I'm fried...

Solve the problem. Use the linear factor theorem to find the polynomial of degree 3 having zeroes x=5, x=1-2i, and x=1+2i. Assume a lead coefficient of 1.

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Mark M. answered • 07/22/17

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If the zeros are 5, 1 - 2i,and 1 + 2i,
Then the factors are (x - 5), (x - (1 - 2i)), and (x - (1 + 2i))
p(x) = (x - 5)(x - 1 + 2i)(x - 1 - 2i)
Can you multiply out?
 
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Victoria V.

Tim,
When you multiply these, multiply the (x-1+2i) and (x-1-2i) first.  You should get a quadratic with NO IMAGINARY TERMS.  They should all cancel out if you have multiplied correctly.  
 
Then, multiply the (x-5) with the new "x2 + bx + c" you obtained above by multiplying the complex conjugates.
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07/22/17

Arturo O.

Tim, 
 
Also keep in mind that if one of the zeros is complex, then its complex conjugate has to be a zero, in order to have real coefficients.  So if the problem asked for a polynomial of degree 3 but only gave you 5 and 1-2i as zeros, you should know that the 3rd root is 1+2i.
 
One you have the 3 zeros x1, x2, and x3, the polynomial comes from expanding
 
f(x) = A(x - x1)(x - x2)(x - x3),
 
where A is real and A ≠ 0.  Since the problem wants a lead coefficient of 1, then
 
A = 1.
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07/22/17

Tim S.

Thank you very much.
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07/22/17

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