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Find all other zeros of P(x)=x^3-6x^2+18x-40 , given that 1+3i is a zero.

Find all other zeros of P(x)=x^3-6x^2+18x-40 , given that 1+3i is a zero.-example-1

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Answer:

1 - 3i, 4

Explanation:

The polynomial has real coefficients, so if it has a complex root, it must have at least two complex roots which are complex conjugates.

Since 1 + 3i is a root, then 1 - 3i is also a root.

Two of the factors of the polynomial are:

x - (1 + 3i) and x - (1 - 3i)

Simplify the factors above:

x - 1 - 3i and x - 1 + 3i

Find their product:

(x - 1 - 3i)(x - 1 + 3i) =

Rewrite them showing we have the product of a sum and a difference.

= [(x - 1) - 3i][(x - 1) + 3i]

Multiply the factors above noticing they are a sum and a difference which follows the pattern (a + b)(a - b) = a² - b².

= (x - 1)² - (3i)²

= x² - 2x + 1 - 9(-1)

= x² - 2x + 10

Now we divide the original polynomial by the product we just found using long division.

x - 4

------------------------------

x² - 2x + 10 | x³ - 6x² + 18x - 40

- x³ - 2x² + 10x

------------------------

-4x² + 8x - 40

- -4x² + 8x - 40

--------------------------

0 + 0 + 0

Now we know that x³ - 6x² + 18x - 40 factors into (x² - 2x + 10)(x - 4).

(x² - 2x + 10)(x - 4) = 0

From x² - 2x + 10, we have roots x = 1 + 3i and x = 1 - 3i.

x - 4 = 0

x = 4

From x - 4, we have root x = 4.

Answer: 1 - 3i, 4

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