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Download "How to Solve Advanced Cubic Equations: Step-by-Step Tutorial"

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  • ruRussian
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00:00:00
hello again everyone in this video
00:00:02
tutorial we are going to solve this
00:00:05
given cubic equation and at the end we
00:00:09
are going to check our answer as well to
00:00:14
make things simple I want to make sure
00:00:16
that you keep in your mind that our this
00:00:19
original equation should look like this
00:00:22
standard form like this one over here
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you can compare it so that means your a
00:00:28
is going to be 2 B is 1 C is negative 13
00:00:33
and B is positive 6 so the very first
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step what we're going to do is we want
00:00:40
to check to see if x equals to 1 and X
00:00:44
equal to negative 1 are the solution to
00:00:48
this equation because those are easy to
00:00:51
find therefore in the very first step
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we're going to be checking whether X
00:00:58
equal to 1 is our solution and here is
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the rule it states that if you add a
00:01:07
plus B plus C plus D and that turns out
00:01:11
to be 0 then one of our solution is
00:01:14
going to be x equal to 1 in our case a
00:01:17
equals to 2 B equals to 1 C equals to
00:01:21
negative 13 and D equal to 6 and if you
00:01:25
add them up this adds up to negative 4
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if you do that 1 which is not a zero so
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that means x equals to 1 is not a
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solution and now in step 2
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we also want to check whether X equal to
00:01:43
negative 1 is a solution and there is a
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rule over here that states that if you
00:01:52
add the alternate coefficients the terms
00:01:58
the coefficients and over here this one
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if you add these coefficients to and
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this one is negative 13 the alternate
00:02:09
terms coefficients so that's going to
00:02:11
give you
00:02:13
negative 11 isn't it and this has a
00:02:17
coefficient 1 if you add this
00:02:20
coefficient to 1 plus 6 becomes 7 are
00:02:26
these numbers same they are not same if
00:02:30
these two numbers are not same that
00:02:35
means this is not X equal to negative 1
00:02:40
this X equal to negative 1 is also
00:02:43
always going to be the solution if these
00:02:46
two numbers must be same so we ruled out
00:02:50
X equal to negative 1 as well so we have
00:02:56
found out that X equal to 1 and X equal
00:03:00
to negative 1 are not our solutions and
00:03:05
in this step now we're going to look for
00:03:08
other possible solutions before we do
00:03:12
that one I want you to look at the
00:03:14
leading coefficient I want you to call
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this one Q and the on the other hand
00:03:20
this number 6 I want you to call P let
00:03:25
me ask you now what is what are your
00:03:31
divisors of P which is 6 so the divisors
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of P are simply 6 R what are the devices
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of 6 positive or negative 1 positive or
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negative 2 positive or negative 3 and
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positive or negative 6
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isn't it and now let's do the positive
00:03:54
the divisors of Q which is true the
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divisors of 2 are going to be a positive
00:04:03
or negative 1 and positive or negative 2
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isn't it
00:04:09
now what I want you to do is I want you
00:04:11
to divide P over Q this is equal to
00:04:16
these numbers on the top like this one
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now what we're going to do is I want you
00:04:23
to simply now we have a P over Q
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you see this thing now we're gonna see
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what are the numbers that we're gonna
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divide this into this one so let's
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divide positive or negative one into all
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these numbers and you divide by one so
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we're gonna get simply positive negative
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one positive negative to positive
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negative 3 and positive negative 6 so
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far so good
00:04:52
now I want you to divide these top
00:04:55
numbers by 2 now positive and negative 2
00:04:57
so 1 over 2 this become positive and
00:05:01
negative 1 over 2 and then 2 over 2 is 1
00:05:05
already we already have a 1 we don't
00:05:07
need to and then finally it's gonna be
00:05:10
positive or negative 3 over 2 and then 2
00:05:16
divided by 6 is 3 which is already here
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so we have these numbers we're gonna see
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if there's any solution among these ones
00:05:26
so the very first thing in the previous
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step we already found out that x equals
00:05:31
to 1 and X equal to negative 1 are not
00:05:35
solution we ruled them out so now let's
00:05:38
try x equals to 2 so now we're going to
00:05:42
try X equal to see whether this is a
00:05:45
solution to this our given cubic
00:05:47
equation for that one we need to use the
00:05:52
synthetic division that's what we're
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gonna do now now let's go ahead and
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start plugging in the numbers right now
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what are the coefficients over here for
00:06:03
this equation 2 over there's no
00:06:07
coefficients can be a 1 this is negative
00:06:09
13 and there's 6 I want you to just
00:06:12
write it down 2 1 negative 13 and 6 and
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the outside number is going to be this X
00:06:21
equal to 2 and let's get started I want
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you to bring down this
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- right up here as it is now I want you
00:06:37
to simply diagonally multiply two times
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two is four add the map one plus four is
00:06:44
five multiply this two times five that's
00:06:48
gonna give you ten so if you add them
00:06:52
that's gonna give you negative three
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multiply this one out so that is gonna
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give you negative six 6 and negative six
00:07:02
they add up to zero and that is our
00:07:08
remainder and now since remainder turned
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out to be equal to zero now I want you
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to use this fact if remainder is 0 then
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this given x equals to 2 is indeed our
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solution so we figured out one of the
00:07:27
solution which is X equal to 2 now what
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next step is I want you to look at these
00:07:33
numbers that we got with synthetic
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division I want you to put X square for
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this one X for this one and and this
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becomes a constant so this could be
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written as 2x squared plus 5x minus 3
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equal to 0 which is a quadratic equation
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you can see that 1 and we can easily
00:08:01
factor it out so what are the possible
00:08:05
factors so the factor factor is going to
00:08:07
be X plus 3 and the other one is going
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to be 2x minus 1 so far so good
00:08:19
now I want you to split them up so X
00:08:23
plus 3 equal to 0 and 2x minus 1 equal
00:08:30
to 0 this is gonna give you X equal to
00:08:34
negative 3 as your one of the solution
00:08:37
and this I want you to bring this
00:08:40
negative 1 on the other side so 2x
00:08:42
become positive on
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by two so X turns out to be 1 over 2 as
00:08:49
your another solution thus our solution
00:08:55
set turns out to be to the one we
00:08:58
already got a two solution and then the
00:09:01
other ones are negative 3 & 1 over 2 now
00:09:05
the in this step finally we want to
00:09:08
check our answer
00:09:09
so as we figured out our solution set
00:09:12
was 2 negative 3 & 1 over 2 I want you
00:09:16
to add all these solutions right up here
00:09:18
I want you to put down 2 plus negative 3
00:09:22
plus 1 over 2 on the left hand side and
00:09:27
ask yourself is this equal to negative B
00:09:32
over a now you might be wondering what
00:09:36
is a and what is B in this original
00:09:39
equation this is our a B is 1 C is
00:09:46
negative 13 and D is 6 so minus B over a
00:09:53
is simply equal to negative 1 over 2 so
00:09:59
that means now if you add and subtract
00:10:03
everything you're gonna see that we're
00:10:05
gonna have a negative 1 over 2 on the
00:10:08
left hand side and on the right hand
00:10:11
side minus B over a it turns out to be
00:10:14
negative 1 over 2 as well isn't it now
00:10:18
ask yourself is this statement true yes
00:10:21
this statement is true so thus X equal
00:10:26
to 2 X equal to negative 3 and X equal
00:10:29
to 1 R 1 over 2 are our solutions and
00:10:33
finally here is the graph of our given
00:10:37
cubic function and you see there are
00:10:39
three X intercepts so these are in other
00:10:43
words they are our solutions thanks for
00:10:46
watching and please subscribe to my
00:10:48
channel for more exciting videos

Description:

Learn how to Solve Advanced Cubic Equations using Synthetic Division. Also learn how to Check your Answer Algebraically and Graphically (Graph of the Cubic Equation is Provided). Step-by-Step Tutorial by PreMath.com

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