Example 2: Multiplying a binomial by a polynomial | Algebra I | Khan Academy

We are multiplying 10a minus
3 by the entire polynomial 5a

squared plus 7a minus 1.

So to do this, we can just
do the distributive property.

We can distribute this
entire polynomial,

this entire trinomial,
times each of these terms.

We could have 5a squared
plus 7a minus 1 times 10a.

And then 5a squared plus 7a
minus 1 times negative 3.

So let's just do that.

So if we have-- so let
me just write it out.

Let me write it this way.

10a times 5a squared
plus 7a minus 1.

That's that right over here.

And then we can have
minus 3 times 5a squared

plus 7a minus 1.

And that is this
distribution right over here.

And then we can simplify it.

10a times 5a squared--
10 times 5 is 50.

a times a squared
is a to the third.

10 times 7 is 70.

a times a is a squared.

10a times negative
1 is negative 10a.

Then we distribute this
negative 3 times all of this.

Negative 3 times 5a squared
is negative 15a squared.

Negative 3 times
7a is negative 21a.

Negative 3 times
negative 1 is positive 3.

And now we can try
to merge like terms.

This is the only a to
the third term here.

So this is 50a to the third.

I'll just rewrite it.

Now we have two a squared terms.

We have 70a squared minus
15, or negative 15a squared.

So we can add these two terms.

70 of something minus
15 of that something

is going to be 55
of that something.

So plus 55a squared.

And then we also
have two a terms.

We have this negative 10a, and
then we have this negative 21a.

So if we go negative 10 minus
21, that is negative 31.

That is negative 31a.

And then finally, we only have
one constant term over here.

We have this positive 3.

So plus 3.

And we are done.