3.09 Polynomial identities and complex numbers

We can extend our knowledge of complex numbers and polynomial identities to find complex factors of polynomials.

With this type of factorization, the key is to introduce the term $i^2=-1$i2=−1 in the given expression whenever required. Let's work through a few examples.

Worked examples

Question 1

Factor the expression $x^2+9$x2+9.

Think: The expression $x^2+9$x2+9 can only be factored if we rewrite the second term using the fact that $i^2=-1$i2=−1. Once it is rewritten, we can apply the difference of two squares polynomial identity.

Do:

$x^2+9$x2+9	$=$=	$x^2-9i^2$x2−9i2	Rewrite the second term using $i^2=-1$i2=−1
	$=$=	$\left(x+3i\right)\left(x-3i\right)$(x+3i)(x−3i)	Factor using the difference of two squares polynomial identity

 

Question 2

Factor the expression $4x^2-12ix-9$4x2−12ix−9.

Think: The middle term has a coefficient of $-12i$−12i. If we rewrite the last term using $i^2=-1$i2=−1, we might be able to factor the expression.

Do:

$4x^2-12ix-9$4x2−12ix−9	$=$=	$4x^2-12i+9i^2$4x2−12i+9i2	Using $i^2=-1$i2=−1, we get $-9=+9i^2$−9=+9i2
	$=$=	$\left(2x-3i\right)^2$(2x−3i)2	Now, factor using the identity $\left(A-B\right)^2=A^2-2AB+B^2$(A−B)2=A2−2AB+B2.

 

Practice questions

Question 3

Question 4

Question 5