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Given that $1-i$ is a zero of $p(x)=x^{3}-6x^{2}+10x-8$ , find the remaining zeroes.

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Complex conjugate theorem

Full solution

Q. Given that

1-i

is a zero of

p(x)=x^{3}-6x^{2}+10x-8

, find the remaining zeroes.

Identify Conjugate Pairs: Since complex roots of polynomials with real coefficients come in conjugate pairs, if $1 - i$ is a zero, then its conjugate, $1 + i$ , must also be a zero of $p(x)$ .
Find Product of Factors: We can use synthetic division or long division to divide the polynomial by $(x - (1 - i))(x - (1 + i))$ to find the remaining zero. First, let's find the product of these two factors.
Perform Polynomial Division: The product of $(x - (1 - i))(x - (1 + i))$ is $(x - 1 + i)(x - 1 - i)$ , which simplifies to $(x - 1)^2 - i^2$ . Since $i^2 = -1$ , this further simplifies to $x^2 - 2x + 1 - (-1)$ , which is $x^2 - 2x + 2$ .
Set Up Division: Now we divide the polynomial $p(x)$ by $x^2 - 2x + 2$ using synthetic division or long division. We set up the division as follows: $\newline$ $x^3 - 6x^2 + 10x - 8 \div (x^2 - 2x + 2)$
Perform Division: Performing the division, we get: $\newline$ $x - 4$ with a remainder of $0$ , since $1 - i$ is indeed a root and its conjugate pair should also be a root, leaving us with a linear factor that gives the remaining real root.
Identify Remaining Zero: The quotient $x - 4$ represents the linear factor of $p(x)$ that gives the remaining real zero. Therefore, the remaining zero is $x = 4$ .

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