How do you write a polynomial in standard form given zeros 8, -14, and 3 + 9i?

Answer 1

#x^4-22x^2-10080=0#

Complex roots occur in conjugate pairs. So, the fourth root is #3-9i#.

If #s_1,s_2, s_3 and s_4# are the sums of the products of the roots, taken 1, 2, 3 and 4, at a time, respectively, the biquadratic equation has the form #x^4-s_1x^3+s_2x^2-s_3x+s_4=0#.

Here, #s_1=0, s_2=-22, s_3=0 and s_4=-10080#.

So, the answer is #x^4-22x^2-10080=0#.

Note that the sum #3+-9i#=0, wherever it occurs in the terms of s-sums.

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Answer 2

Theodore Amidon

To write a polynomial in standard form given zeros 8, -14, and 3 + 9i, you would start by using the zeros to construct the factors of the polynomial. Since 3 + 9i is a complex zero, its conjugate 3 - 9i is also a zero. Then, you would multiply these factors together to get the polynomial expression. Finally, you would expand and simplify the expression to put it in standard form.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.