Question

Find a polynomial with real coefficients that has 2, 3 + i, and 3 - 2i as zeros.

Answer 1

Answer:

Standard form: $x^5-14x^4+83x^3-256x^2+406x-260$

Factored form with complex numbers: $(x-2)(x-(3+i))(x-(3-i))(x-(3-2i))(x-(3+2i))$

Factored form without complex numbers: $(x-2)(x^2-6x+10)(x^2-6x+13)$

Step-by-step explanation:

If the polynomial has real coefficients and a+bi is a zero, then a-bi zero.

This means the following:

1) Since 3+i is a zero, then 3-i is a zero.

2) Since 3-2i is a zero, then 3+2i is a zero.

So we have the following zeros:

1) x=2

2) x=3+i

3) x=3-i

4) x=3-2i

5) x=3+2i

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

This implies the following:

1) x=2 is a zero => x-2 is a factor

2) x=3+i is a zero => x-(3+i) is a factor

3) x=3-i is a zero => x-(3-i) is a factor

4) x=3-2i is a zero => x-(3-2i) is a factor

5) x=3+2i is a zero => x-(3+2i) is a factor

Let's put our factors together:

$(x-2)(x-(3+i))(x-(3-i))(x-(3-2i))(x-(3+2i))$

Let's find the standard form for this polynomial.

This will require us to multiply the above out and simplify by combining any like terms.

Before we begin this process, I would rather have a quick way to multiply the factors that contain the congregate pair zeros.

$(x-(m+ni))(x-(m-ni))$

I'm going to use foil.

First: $x(x)=x^2$

Outer: $x(-(m-ni))=-(m-ni)x$

Inner: $-(m+ni)(x)=-(m+ni)x$

Last: $-(m+ni)(-(m-ni))=(m+ni)(m-ni)=m^2-n^2i^2=m^2-n^2(-1)=m^2+n^2$

(Note: $(a-b)(a+b)=a^2-b^2$; When multiplying conjugates, you just have to multiply the first terms of each and the last terms of each.)

------------------------------------Let's combine these terms:

$x^2-(m-ni)x-(m+ni)x+m^2+n^2$

Distribute:

$x^2-mx+nix-mx-nix+m^2+n^2$

Combine like terms:

$x^2-2mx+m^2+n^2$

So the formula we will be using on the factors that contain conjugate pair zeros is:

$(x-(m+ni))(x-(m-ni))=x^2-2mx+m^2+n^2$.

$(x-(3+i))(x-(3-i))=x^2-2(3)x+3^2+1^2$

$(x-(3+i))(x-(3-i))=x^2-6x+9+1$

$(x-(3+i))(x-(3-i))=x^2-6x+10$

$(x-(3+2i))(x-(3-2i))=x^2-2(3)x+3^2+2^2$

$(x-(3+2i))(x-(3-2i))=x^2-6x+9+4$

$(x-(3+2i))(x-(3-2i))=x^2-6x+13$

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

So this is what we have now:

$(x-2)(x^2-6x+10)(x^2-6x+13)$

I'm going to multiply the last two factors:

$(x^2-6x+10)(x^2-6x+13)$

What we are going to do is multiply the first term in the first ( ) to every term in the second ( ).

We are also going to do the same for the second term in the first ( ).

Then the third term in the first ( ).

$x^2(x^2)=x^4$

$x^2(-6x)=-6x^3$

$x^2(13)=13x^2$

$-6x(x^2)=-6x^3$

$-6x(-6x)=36x^2$

$-6x(13)=-78x$

$10(x^2)=10x^2$

$10(-6x)=-60x$

$10(13)=130$

--------------------------------Combine like terms:

$x^4-12x^3+59x^2-138x+130$

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

So now we have

$(x-2)(x^4-12x^3+59x^2-138x+130$

We are almost done.

We are going to multiply the first term in the first ( ) to every term in the second ( ).

We are going to multiply the second term in the first ( ) to every term in the second ( ).

$x(x^4)=x^5$

$x(-12x^3)=-12x^4$

$x(59x^2)=59x^3$

$x(-138x)=-138x^2$

$x(130)=130x$

$-2(x^4)=-2x^4$

$-2(-12x^3)=24x^3$

$-2(59x^2)=-118x^2$

$-2(-138x)=276x$

$-2(130)=-260$

----------------------------------------Combine like terms:

$x^5-14x^4+83x^3-256x^2+406x-260$