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

Question

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

Bytelearn · Accepted Answer

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:  x^3 - 6x^2 + 10x - 8 ÷ (x^2 - 2x + 2)Perform Division: Performing the division, we get:  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.

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

Full solution

More problems from Complex conjugate theorem

Related topics

Given that 1−i1-i1−i is a zero of p(x)=x3−6x2+10x−8p(x)=x^{3}-6x^{2}+10x-8p(x)=x3−6x2+10x−8, find the remaining zeroes.

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