Writing an equation in standard form using the zeros x=(-2-√5)/3 and x=(3/4) + i√7

Question

Here is the question:

Writing an equation in standard form for a polynomial with zeros x=(-2-√5)/3 and x=(3/4) + i√7.

The function must give those 2 zeros including the imaginary one.

Please show work so I understand the function better.

Answer 1

All irrational roots come in pairs and all imaginary roots come in pairs. So this function would have at least four zeros. It will be easier to work with each pair of zeros separately to get two second degree trinomials and then multiply those together at the end to get a fourth degree polynomial.

Let’s start with the irrational root x= (-2-√5)/3. If we think about the quadratic formula, there is ± before the radical. Since the root we are given is -√5, there has to be a pair that includes +√5. So we know that there is a second zero at x= (-2+√5)/3. Working with radicals and fractions can be difficult, so let’s manipulate the zero to eliminate as much of that as we can.

1. start with x=(-2-√5)/3. We want this in factor form. First multiply by 3 on both sides so we get 3x= -2-√5
2. then move everything over so the right side is equal to zero. So we get 3x+2+√5 as a factor
3. if we do the same for the irrational pair zero, we’ll get 3x+2-√5 as a factor
4. notice that the only difference in the factors is the -√5 and +√5. Lets group the factors so they’re easier to multiply and that one difference stands out. So our two factors are (3x+2)+√5 and (3x+2)-√5.
5. This should look similar to the factors of a difference of squares polynomial. When we FOIL factors from a difference of squares polynomial (ie (x+4)(x-4), the outer and inner terms always cancel out (ie -4x and 4x). So we really only need to multiply the first terms and last terms. For this example, that means that we have to multiply the first terms, (3x+2)(3x+2) and the last terms, (+√5)(-√5). This gives us 9x²+12x+4-5 which simplifies to 9x²+12x-1.

Now lets look at the complex zeros. If one zero is at x= 3/4 +i√ 7, then there is another zero at its complex conjugate: x= 3/4 -i√ 7. Lets manipulate these similar to how we manipulated the irrational roots.

1. Start with x= 3/4 +i√ 7. We want this in factor form. First multiply by 4 on both sides so we get 4x= 3+4i√7.
2. then move everything over so the right side is equal to zero. So we get 4x-3-4i√7 as a factor
3. if we do the same for the complex conjugate pair zero, we’ll get 4x-3+4i√7.
4. notice that, similar to the irrational factors, the only difference in the factors is the -4i√7 and +4i√7. Lets group the factors so they’re easier to multiply and that one difference stands out. So our two factors are (4x-3)-4i√7 and (4x-3)+4i√7.
5. similar to the irrational factors, when we multiply, we only have to multiply the first terms, (4x-3)(4x-3) and the last terms (-4i√7)(+4i√7). This gives us 16x²-24x+9-16i²(7). Remember to substitute -1 for i². So -16i²(7) equals -16(-1)(7) which equals +112. So our polynomial simplifies to 16x²-24x+121.

Lastly we need to multiply our two quadratic polynomials to get our fourth degree polynomial equation. You can use distribution or the box method for this step. Just be very careful with the exponents. After multiplying, our final answer is 144x⁴-24x³+785x²+1476x-121.

This can be checked on the calculator. Type this polynomial into Y=. Then click 2nd,TRACE,value (option 1) and check (-2-√5)/3 and (-2+√5)/3. Both values will give y=0 as an answer, which means they are in fact zeros.

Answer 2

Often times these types of problems say to make sure the polynomial has integer coefficients. Since that is not mentioned, I'll assume that is not a requirement.

A critical thing to know about complex roots (roots with imaginary numbers) is that they NEVER appear by themselves. They ALWAYS appear as conjugate pairs. A conjugate has the same numbers but is seperated by the opposite sign. For the complex number 2 + 3i, the number 2 - 3i is its conjugate. So, in this case, they tell you that one of the zeros is x = 3/4 + i√7 meaning that there MUST also be a zero that is x = 3/4 - i√7

Wityh this information, you know have 3 zeros that you can use to form the polynomial. It will be the product of [x - (-2-√5)/3][x - (3/4 + i√7)]{ x - (3/4 - i√7)] which simplifies to:

(x + 2/3 + √5/3)(x - 3/4 - i√7)(x - 3/4 + i√7)

You can multiply these out however you like to do so.

I prefer to first multiply the complex numbers. Since they are conjugate pairs, their product will simplify down nicely. I also prefer using a rectangle like this:

Several of the elements cancel out and, since i² = -1, we can simplify -7i² to be +7 so the whole thing turns out to be:

x² - 1.5x + 7 + 9/16 or x² - 3/2x + 121/16

Then we can multiply this by (x + 2/3 + √5/3) using another rectangle:

This simplifies to x³ - 5/6x² + √5/3x² + 105/16x - √5/2x + 121/24 + 121√5/48

To officially be in "standard form" I would probably combine the coefficients together like this:

x³ - (5/6 - √5/3)x² + (105/16 - √5/2)x + 121/24(1 + √5/2)