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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.


$x^2+9$x2+9 $=$= $x^2-9i^2$x29i2

Rewrite the second term using $i^2=-1$i2=1

  $=$= $\left(x+3i\right)\left(x-3i\right)$(x+3i)(x3i)

Factor using the difference of two squares polynomial identity

Question 2

Factor the expression $4x^2-12ix-9$4x212ix9.

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.


$4x^2-12ix-9$4x212ix9 $=$= $4x^2-12i+9i^2$4x212i+9i2

Using $i^2=-1$i2=1, we get $-9=+9i^2$9=+9i2

  $=$= $\left(2x-3i\right)^2$(2x3i)2

Now, factor using the identity $\left(A-B\right)^2=A^2-2AB+B^2$(AB)2=A22AB+B2.


Practice questions

Question 3

Complete the factoring by filling in the empty box.

  1. $9ix-54=9i\left(\editable{}\right)$9ix54=9i()

Question 4

Factor the expression $3x^2+108$3x2+108. Leave your answer in terms of $i$i.

Question 5

Factor the expression $x^2+12ix-36$x2+12ix36.



Extend polynomial identities to the complex numbers.

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