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.
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 |
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. |
Complete the factoring by filling in the empty box.
$9ix-54=9i\left(\editable{}\right)$9ix−54=9i()
Factor the expression $3x^2+108$3x2+108. Leave your answer in terms of $i$i.
Factor the expression $x^2+12ix-36$x2+12ix−36.