Solving second-degree polynomials. By
second-degree, we mean that the highest
power is a two. It's also true that there
are usually two solutions. Here's a
second-degree polynomial. Watch out for this. If there's only an x
squared term and there's no other terms
with variables, then just get that by
itself and take the square root. So I'm
going to add five to both sides. I have
3x squared left and 21 on that side.
These are gone. Divide by 3 on both
sides to get x squared by itself. So x
squared is 7. When we take the square
root to get x squared by itself,
remember this is plus or minus the
square root of 7. And I can't simplify
that. These are my two solutions: X is
square root of 7 or negative square root
of 7. This one's different.
Now, in addition to x squared, we also
have an X term. You cannot get x squared
by itself and take the square root
because you'll be stuck with an X on the
other side of the equation. And, you can't
add or subtract these terms together
because they're not like terms, so we
need a new strategy. The new strategy is
going to be using factoring, so let's
review how to factor a trinomial like
this. In another video, we talked about
using the box method. There's other
methods too and they all come out the
same answer; as long as you know how to
factor it's fine. But I'm going to use
the box method. So remember I've got x
squared in one box and 21 in another.
And in the other two boxes, I need two x
terms that add up to negative 10x.
They're also made of the factors of 21.
So for factors of 21, I've got 1 times 21 and 3 times 7. The only way I'm going to get a
pair to add up to negative 10 and
multiply together to give me positive 21
is if they're both negative, so look at
it this way. Which of these pairs adds up
to negative 10? Clearly it's this one. So in these two
boxes, we put negative 3x and negative 7x. The order does not matter. Now factor up
the columns and across the rows. So up
here, I factor out a common factor. The
only common factor is X, and I pull out a
positive x because the first term is
positive, the one on top. Next I have
something negative because the top term
is negative. The common factor they both
have is 3, so that's what I factored out
of that one. Now we factor across this
way. Since x squared is positive, I'm
pulling out a positive common factor,
which is X. And now this way. Since
negative 7x is negative, I'm pulling out
a negative factor. And the common factor
of negative 7x and 21 is 7. So now I have
my two binomial factors: X minus 3 and X
minus 7. And if you multiply those back
together you would get this trinomial. So
these are the factors. So let's see how
that helps us solve that equation. Well
now I'm going to use something new
called the zero product property. Here's
what the property states: if you have a
number A (any number) times some other
number, let's call it B, and the product
of those two numbers is zero, we know for
sure that either A could be zero or B could
be zero. And those are both solutions to
that equation. Because if A is zero, it
doesn't matter what B is,
but if B is zero, it doesn't matter what A
is. It would wipe it out, right? So we're
going to use that property to solve this.
Either this or this must be equal to zero,
so solve both. Either X minus 3 equals zero or X minus 7 equals zero. This one comes out
to be x equals 3, and this one is x
equals 7. These are both solutions to
this problem. Let's try plugging those
numbers back into the original
polynomial to see if they work.
If X was 3, I would get 3 squared minus
10 times 3 plus 21. Does that equal zero?
Let's see. 9 minus 30 plus 21. That is
negative 21, so that's zero. Okay so 3 is
definitely a solution. Let's try 7. 7
squared minus 10 times 7 plus 21.
Ok that's 49 minus 70 plus 21. Does that
equal zero? Well this is negative 21,
so again that's going to equal zero. And
those are both solutions. Make sure that
you circle or box in your solution, and
make sure that you check your solutions
to make sure that they work. You could
have made an error and it would show up
if you plug it back in, it won't work. And
that should tell you that you need to do some more work. Let's try another one. Now
this one has an x squared and an X term,
so I do have to use factoring. But I
can't use the zero product property
unless it's equal to zero, so before you
start, you need to subtract 10 from both
sides to get it equal to zero. So that
turns into x squared minus 3x minus 10. I
can't really combine those, they're not
like terms right? Now that it's equal to
zero, we factor it. Remember how to do that.
Ok this time our product is negative 10,
so keep that in mind, one of the factors
has to be negative and the other has to
be positive. Since the middle term is
negative, that means the larger factor
needs to be negative, so we do this. Now
which of those makes a sum that matches our middle term, that negative 3? It's
this one right? That makes negative 3. So
those are the factors. Those are our X
terms, so we put 2x and negative 5x and then factor out common
factors. Both of those, 2x and negative 10, have a 2 in common. And since the top
term was positive, that's going to be
positive 2. And then we go this way, so
the common factor in the top row is X,
and in the bottom row it's going to be
negative 5. Our two factors are X plus 2 and X minus 5, and that's still equal to zero.
Now use the zero product property. So
either X plus 2 equals zero, or X minus 5
equals zero. That means X has to be negative 2 or 5. These are both solutions to the
problem. Let's check them. If X was
negative 2 in the original problem, that
would be negative 2 squared minus 3
times negative 2, should be equal to 10.
Let's see if it really is. 4 plus 6, yep
that's 10. Now let's plug in 5. 5 squared
minus 3 times 5 equals 10. That's 25
minus 15. That's definitely 10. Alright
so there's two kinds of problems you
need to watch out for. If you see an x
squared and an X, you need to use
factoring, you need to set it equal to
zero. Okay if it just has x squared and
no other variables, then get the x
squared by itself and take the square
roots. And those are your new methods for
solving second-degree equations.
