### Video Transcript

Complete the blanks in this
equation. Something multiplied by something
minus ~~π~~ [ππ] to the fourth equals three π to the forth
π to the eighth minus π squared π to the eighth.

The distributive property is being
used here. But itβs our job to figure out what
exactly is being distributed. Our missing value here was
distributed across this subtraction problem to produce three π to the fourth π to
the eighth minus π squared π to the eighth.

Letβs focus our energy on these two
pieces. How would we go from π times π to
the fourth to π squared π to the eighth? To figure out what was being
multiplied here, we can use division. We divide π squared π to the
eighth by π times π to the forth. When we do that, our π squared
becomes π to the first power, and our π to the eighth becomes π to the fourth
power. This is the value weβll plug in
here, π times π to the fourth. Itβs the value being distributed
across this problem.

Now we need to focus on our last
missing term. We need to know what π times π to
the fourth was multiplied by to produce three π to the fourth π to the eighth. We can use the same process
here. We can use division to find out
what was being multiplied. Weβll divide three π to the fourth
π to the eighth by ππ to the fourth. The three canβt be divided, so we
keep that. Our π to the fourth becomes π
cubed, and our π to the eighth becomes π to the fourth. This means the final missing value
would be three π cubed π to the fourth.

Our two missing pieces were π
times π to the fourth, three times π cubed times π to the fourth.