Hi, I need help. I don't really get what am I supposed to do

Question

Given the zeros in each problem, use factors to find the original polynomial

1. -1,1+3i'
2. -1/4,1+√6
3. 1+√3,-3+√5

THEY ARE WRITTEN CORRECTLY. PLEASE HELP ME WITH ALL THE PROBLEMS THANK YOU :)

Answer 1

Hi Kim, hope this helps you.

Problem #1: You are given 2 zeros: -1 and 1 + 3i

due to the Complex Conjugate Root Theorem, you also have the zero: 1 - 3i

from these 3 zeros, you can get the factors of the polynomial

the zero x = -1 gives you the factor x + 1 = 0

the zero x = 1 + 3i gives you the factor x - 1 - 3i = 0

the zero x = 1 - 3i gives you the factor x - 1 + 3i = 0

if you multiply these 3 factors, their product equals to zero

(x + 1)(x - 1 - 3i)(x - 1 + 3i) = 0

(x + 1)(x² - x + 3ix - x + 1 - 3i - 3ix +3i - 9i²) = 0

(x + 1)(x² - 2x + 1 - 9(-1)) = 0 where i² = -1

(x + 1)(x² - 2x +10) = 0

x³ - 2x² + 10x + x² - 2x + 10 = 0

x³ - x² + 8x + 10 = 0

so the polynomial you are looking for is p(x) = x³ - x² + 8x + 10

Problem #2: you are given 2 zeros: -1/4 and 1 + √6

due to the Irrational Conjugate Theorem, you also have the zero 1 - √6

from these 3 zeros you can get the factors of the polynomial

the zero x = -1/4 gives you the factor x + 1/4 = 0 but we will use 4x + 1 = 0 to avoid fractions

the zero x = 1 + √6 gives you the factor x - 1 - √6 = 0

the zero x = 1 - √6 gives you the factor x - 1 + √6 = 0

if you multiply these 3 factors, their product equals to zero

(4x + 1)(x - 1 - √6)(x - 1 + √6) = 0

(4x + 1)(x² - x + x√6 - x + 1 - √6 - x√6 + √6 - √36) = 0

(4x + 1)(x² - 2x + 1 - 6) = 0

(4x + 1)(x² - 2x - 5) = 0

4x³ - 8x² - 20x + x² - 2x - 5 = 0

4³ - 7x² - 22x - 5 = 0

so the polynomial you are looking for is p(x) = 4x³ - 7x² - 22x - 5

Problem #3: you are given the zeros 1 + √3 and -3 + √5

due to the Irrational Conjugate Theorem you also have the zeros 1 - √3 and -3 - √5

from these 4 zeros you can get the factors of the polynomial you are looking for

the zero x = 1 + √3 gives you the factor x - 1 - √3 = 0

the zero x = 1 - √3 gives you the factor x - 1 + √3 = 0

the zero x = -3 + √5 gives you the factor x + 3 - √5 = 0

the zero x = -3 - √5 gives you the factor x + 3 + √5 = 0

if you multiply these 4 factors, their product equals to zero

(x - 1 - √3)(x - 1 + √3)(x + 3 - √5)(x + 3 + √5) = 0

hint: multiply the first 2 factors, then multiply the last 2 factors

(x² - x + √3x - x + 1 - √3 - √3x + √3 - 3)(x² + 3x + √5x + 3x + 9 + 3√5 - √5x - 3√5 - 5) = 0

(x² - 2x - 2)( x² + 6x + 4) = 0

x⁴ + 6x³ + 4x² - 2x³ - 12x² - 8x -2x² - 12x - 8 = 0

x⁴ + 4x³ - 10x² - 20x - 8 = 0

so the polynomial you are looking for is p(x) = x⁴ + 4x³ - 10x² - 20x - 8

Well, I hope this helps you.

Answer 2

If a is a zero, then x - a is a factor of the polynomial.

complex roots come in conjugate pairs

irrational roots comes in conjugate pairs.

Use these to answer.