NZ Level 8 (NZC) Level 3 (NCEA) [In development] Solutions to Polynomials
Lesson

## Polynomials

A polynomial in $z$z of degree $n$n has the form $P\left(z\right)=a_0+a_1z+a_2z^2+....+a_nz^n$P(z)=a0+a1z+a2z2+....+anzn, where the coefficients $a_0,a_1,a_2,...a_n$a0,a1,a2,...an are either purely real or complex.

A common convention is to use $z$z as the variable rather than $x$x whenever we are interested in solutions to the polynomial equation  $P\left(z\right)=0$P(z)=0 over the complex number field.

## The conjugate root theorem

In the specific instance where the coefficients are real, the complex solutions to $P\left(z\right)=0$P(z)=0, should there be any, will always occur in conjugate pairs

This beautiful result is known as the Conjugate Root Theorem. Specifically, it states that if $a+bi$a+bi is a root of $P\left(z\right)=0$P(z)=0, then so also is its conjugate $a-bi$abi

Without going into a full proof, we can demonstrate with an example why this must be case. To do this we first need to state a few results on conjugates.

### Preliminary conjugate results

With a little algebra, we can easily show that the conjugate of a sum of any two complex numbers is the same as the sum of the conjugates of each complex number.

That is to say, we can show that$\overline{\left(a+bi\right)+\left(c+di\right)}=\overline{a+bi}+\overline{c+di}$(a+bi)+(c+di)=a+bi+c+di with the following argument:

 $LHS$LHS $=$= $\overline{\left(a+bi\right)+\left(c+di\right)}$(a+bi)+(c+di) $=$= $\overline{\left(a+c\right)+\left(b+d\right)i}$(a+c)+(b+d)i $=$= $\left(a+c\right)-\left(b+d\right)i$(a+c)−(b+d)i $=$= $\left(a-bi\right)+\left(c-di\right)$(a−bi)+(c−di) $=$= $\overline{a+bi}+\overline{c+di}$a+bi+c+di $=$= $RHS$RHS

With similar arguments, we can also show that:

• the conjugate of the difference of two complex numbers is the difference in the conjugates
• the conjugate of the product of two complex numbers is the product in the conjugates  .

So for example we know that:

$\overline{\left(1+3i\right)-\left(2-i\right)}=\overline{1+3i}-\overline{2-i}$(1+3i)(2i)=1+3i2i

and that:

$\overline{\left(1+3i\right)\times\left(2-i\right)}=\overline{1+3i}\times\overline{2-i}$(1+3i)×(2i)=1+3i×2i.

Note that this also means that $\overline{2z}=\overline{2}\times\overline{z}=2\overline{z}$2z=2×z=2z since the conjugate of the real number $2$2, like the conjugate of all reals, is itself.

It also means that, for integer powers $n$n,  $\overline{z^n}=\left(\overline{z}\right)^n$zn=(z)n. So, for example $\overline{z^2}=\left(\overline{z}\right)^2$z2=(z)2

### Pointing the way to a proof with a simple example

Suppose we know that the complex number $2+2i$2+2i is a root of the polynomial  $P\left(z\right)=z^2+4z+8$P(z)=z2+4z+8

We ask ourselves is $z=\overline{2+2i}$z=2+2i also a root?

If we substitute $\overline{2+2i}$2+2i into $P\left(z\right)$P(z) we have:

 $P\left(\overline{2+2i}\right)$P(2+2i) $=$= $\left(\overline{2+2i}\right)^2+4\left(\overline{2+2i}\right)+8$(2+2i)2+4(2+2i)+8 $=$= $\overline{\left(2+2i\right)^2}+\overline{4\left(2+2i\right)}+\overline{8}$(2+2i)2+4(2+2i)+8 $=$= $\overline{\left(2+2i\right)^2+4\left(2+2i\right)+8}$(2+2i)2+4(2+2i)+8 $=$= $\overline{0}$0 $=$= $0$0

In other words, the conjugate $2-2i$22i is also a root of $P\left(z\right)$P(z)

It's not hard to see how this result can be extended to any polynomial $P\left(z\right)$P(z) with real coefficients. Can you see why the argument will break down if any of the coefficients of $P\left(z\right)$P(z) are complex?

What this all means is, that if a polynomial function has real valued coefficients, then its complex roots, if there are any, exist as conjugate pairs.

### Using the conjugate root theorem

The conjugate root theorem becomes useful even when finding the roots of polynomial functions with real domains.

$P\left(x\right)$P(x) of degree $n$n has either $n$n real roots, or else $n-2$n2 or $n-4$n4 or $n-6$n6, etc real roots, because any complex roots it has will always exist in conjugate pairs.

We discuss a few examples to illustrate the point.

##### Example 1

Consider the monic cubic function $y=f\left(x\right)$y=f(x) that has only one (and no other) real root at $x=a$x=a.

Then we know that it must have the form $y=\left(x-a\right)\left(x^2+qx+r\right)$y=(xa)(x2+qx+r) where the contained quadratic polynomial expression is irreducible  and hence has no real zeros itself.

However, if we admit complex numbers as candidates for zeros, we know that there are two other zeros that exist as a complex conjugate pair.

In addition, suppose we also know that this function has a complex root of $x=2+i$x=2+i and a particular function value given by $f\left(0\right)=35$f(0)=35.

With this new information, we immediately know two more things:

1. It must have another complex root of $x=2-i$x=2i. In other words, the function is expressible as $f\left(x\right)=\left(x-a\right)\left(x-2+i\right)\left(x-2-i\right)$f(x)=(xa)(x2+i)(x2i).
2. By multiplying out these last two factors, we see that, equivalently,  $f\left(x\right)$f(x) is given by $f\left(x\right)=\left(x-a\right)\left(x^2-4x+5\right)$f(x)=(xa)(x24x+5). Since $f\left(0\right)=35$f(0)=35, this implies that $-5a=35$5a=35 and $a=-7$a=7, and so the three roots are $2+i$2+i, $2-i$2i, and $-7$7.

##### Example 2

How many real roots might a degree $6$6 polynomial equation with real coefficients have if it is know to have the complex roots $1-i$1i, $2-i$2i, and $3-i$3i?

The clear answer is none, because, by the conjugate root theorem, the other three roots are $1+i$1+i, $2+i$2+i and $3+i$3+i

##### Example 3

The polynomial $P\left(x\right)=x^3-x^2-20x+50$P(x)=x3x220x+50 has a zero of $-5$5. Find the other zeros.

Here is one way we can proceed:

We can immediately write that $x^3-x^2-20x+50=\left(x+5\right)\left(x^2+px+q\right)$x3x220x+50=(x+5)(x2+px+q) and from equating the constant terms on both sides, we know that $5q=50$5q=50, from which $q=10$q=10.

Equating the terms in $x$x on both sides reveals that $q+5p=-20$q+5p=20 and since $q=10$q=10, we have that $p=-6$p=6.

Thus the quadratic factor becomes $x^2-6x+10$x26x+10. This could also have been deduced directly using polynomial division.

A quick calculation shows that the discriminant of the quadratic is $-4$4, and so $P\left(x\right)$P(x) must only have exactly one real zero.

If we wish to find the complex roots, we could solve the equation   $x^2-6x+10=0$x26x+10=0. We can do this by completing squares:

 $x^2-6x+10$x2−6x+10 $=$= 0 $x^2-6x+9+1$x2−6x+9+1 $=$= 0 $\left(x-3\right)^2-i^2$(x−3)2−i2 $=$= 0 $\left(x-3-i\right)\left(x-3+i\right)$(x−3−i)(x−3+i) $=$= 0 $\therefore$∴    $x$x $=$= $3\pm i$3±i

#### Worked Examples

##### Question 1

The polynomial $P\left(x\right)=x^3-7x^2+17x-15$P(x)=x37x2+17x15 has a zero at $x=3$x=3. Solve for the other zeros of $P\left(x\right)$P(x).

##### Question 2

The polynomial $P\left(x\right)=x^4+8x^2-9$P(x)=x4+8x29 has a zero at $x=3i$x=3i. Solve for the real zeros of $P\left(x\right)$P(x).

### Outcomes

#### M8-9

Manipulate complex numbers and present them graphically

#### 91577

Apply the algebra of complex numbers in solving problems