31.0k views
0 votes
Completely factor x⁴ + 3x³ + 5x² + 27x - 36 into linear functions given that -3i is a zero?

User Drogon
by
7.6k points

1 Answer

4 votes

Final answer:

To factor the given polynomial with -3i as a zero, we consider that complex zeros come in conjugate pairs, so +3i is also a zero. After dividing the polynomial by (x + 3i)(x - 3i), we determine the quadratic factor and potentially apply the quadratic formula to find the remaining zeros, thus fully factoring the polynomial.

Step-by-step explanation:

To completely factor the polynomial x⁴ + 3x³ + 5x² + 27x - 36 into linear functions given that -3i is a zero, we must use the fact that complex zeros of polynomials with real coefficients come in conjugate pairs. This means that if -3i is a zero, so is its conjugate 3i. Therefore, the polynomial can be written as (x + 3i)(x - 3i) times a quadratic factor.

First, we'll find the quadratic factor by dividing the original polynomial by the product of the factors corresponding to the known zeros:

(x⁴ + 3x³ + 5x² + 27x - 36) ÷ ((x + 3i)(x - 3i))

The product (x + 3i)(x - 3i) simplifies to x² + 9, which is a difference of squares. Now, we can perform polynomial division or use a synthetic division approach to determine the quadratic factor. After factoring the quadratic expression, if possible, we can then express the original polynomial as a product of linear factors.

Let's assume the division gives us a quadratic factor, which we can represent as ax² + bx + c. We can then factor this quadratic expression further using the quadratic formula if it doesn't factor easily:

For the equation ax² + bx + c = 0,

The solution for x is given by:

x = (-b ± √(b² - 4ac)) / (2a)

Applying this to our quadratic factor yields the remaining zeros. Combining these with the known complex zeros gives us the fully factored form of the original polynomial.

User Euthyphro
by
7.6k points