How do you write a polynomial in standard form given the zeros x=7, -13, and 3+2i?

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Answer 1 · 2016-10-02T11:27:30

$"The Reqd. Poly."=k(x^4-114x^2+624x-1183), k in CC-{0}.$

Let $p(x)$ be the reqd. poly.

The Zeroes of $p(x)" are "7,-13, and, 3+2i$ .

We know that the complex zero of a poly. occurs in conjugate pairs.

So, the conjugate of $3+2i, i.e, 3-2i$ is also a zero of $p(x).$

In all, thus, we have $4$ zeroes of $p(x)$ , meaning that $p(x)$ must

be of degree $4$ , having factors, derived using the zeroes,

$(x-7), (x+13), (x-3-2i), & (x-3+2i).$

We conclude that,

$p(x)=k(x-7)(x+13)(x-3-2i)(x-3+2i), k in CC-{0}$

$=k(x^2+6x-91){(x-3)^2-(2i)^2}$

$=k(x^2+6x-91){x^2-6x+9-4i^2}$

$=k(x^2+6x-91){x^2-6x+9-4(-1)}$

$=k(x^2+6x-91)(x^2-6x+13)$

$=-k{6x+(x^2-91)}{6x-(x^2+13)}$

$=-k{(6x)^2+6x(x^2-91-x^2-13)-(x^2-91)(x^2+13)}$

$=-k{36x^2-624x-(x^4-78x^2-1183)}$

$-k(114x^2-624x-x^4+1183)$

$:. p(x)=k(x^4-114x^2+624x-1183), k in CC-{0}.$