Inecuaciones de grado mayor a 2 | Inecuación de cuarto grado división sintética | Ejemplo 8

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Hello, in this video we are going to solve the

following fourth-degree equation.

To solve it, remember that

we need one side of the equation to be

the number 0 and the other side to have

first-degree polynomials

multiplying or dividing, and we don't

have that.

Therefore, we need to factor this

expression, but there is no

factorization case that we can apply.

Therefore, we are going to apply synthetic division.

Remember that for this, we must have

the polynomial organized in

descending order, that is, from the highest

power of the letter x to the lowest.

So, in this case, we already have it

organized. So we start with x to the

4th power, and the number that accompanies it is the

number 1.

x to the 3rd power is accompanied by 2. It should continue with

x to the 2nd power.

If x to the 2nd power were not there, I would have to put

a 0. It is only in the present case. If it

were not there, perhaps 1, or if it were not there, x

to the 3rd

power. All the smaller ones, perhaps 4, are fine. And in case they are not there, we

put 0.

Here appears the one that accompanies it, perhaps

2, is negative 7,

the one that accompanies it, perhaps the 18

and the constant term is

positive 12. These are the values ​​we'll

use to perform the division.

Now we look at which number we can

divide by.

Remember, we focus on the

constant term. The candidates for

an exact division are the divisors of

that number. 12 is divided by plus

and minus 1,

plus and minus 2, plus and

minus 3, 4 and -4, plus 6 and -6,

and plus 12 and minus

12. All of

these are the divisors of 12.

All these positive and

negative numbers. As you can see, we have several

options. If you want, you can choose them

randomly. I recommend doing it

in order. I'm going to try with a positive one.

Now remember how synthetic division is done: bring

down the first number and

start multiplying one by one. This gives me

one, and we place it below the

next number. Here we perform

addition or subtraction depending on the

signs. Two positive numbers and one positive number

gives us three positive numbers. We repeat the

process: one times three equals three, and minus

seven plus three gives me minus four.

1 x minus 4 = 4 - 8 -4 - 12 1 times negative

12 is negative 12 and 12 minus 12 gives me 0. In

this way we already found the first

exact division and it gives us when we have

the value 1. Remember how we found the

divisor polynomial, we say that x is equal

to this number and what we do is set it

equal to 0. x minus 1 is equal to zero, that is,

we divide this polynomial by

x and 1 and it gave us an exact division. What

would the result be? Then remember

that synthetic division lowered the

degree of the polynomial. If here it is perhaps

the 4th, the result of being x to the 3rd power, then it

would be 1 x to the 3rd power, which would be the same as

removing the cube alone, since there is no

need to put the 1. Positive 3

accompanied by x squared, which is what

comes after x, next to it is

perhaps 1. So negative 4 would accompany

x to the 1st power and then the independent term,

so it is the result of the division.

x, the divisor polynomial,

should give us the original polynomial.

So we already factored that

polynomial once and I am left with expressed as a

third-degree polynomial multiplied

by a first-degree polynomial, but as we

can see, we still need to

lower the degree. Therefore, I'm going to

apply synthetic division again to

factor this expression. We take

the coefficients, which are the values ​​that

appear here, and I'm going to put them in to

do the synthetic division again.

We're going to continue dividing by the

other options we had, since we're still

with the number minus 2, and

so I already divided by plus 1. Now I'm going

to try with minus 1. We

start our synthetic division,

remember, by bringing down the first number and we

start multiplying. Minus 1 by

1 gives us minus 1 and 3. Minus 1

gives us 2.

Now, minus 1 x 2 gives us minus 2 and minus

4. Minus 2 gives us minus 6. Now, minus 1

x minus 6 gives me positive 6 and minus

12. 6 minus 6. As we can see, this

division is not exact, therefore, minus

1 didn't work for us. So we're going to

try with another number.

We're going to try with positive 2.

We do the same procedure: bring down 1

and 2 times 1 would give me 2, 3 and positive 2 would give me

5 gives positive 2 times 5 is 10 and minus 4 +

10 gives me that device 2 times 6 gives me

12 and minus 12 12 is zero we find

another root of this polynomial which is the

number 2 what would be the divisor polynomial

then x is equal to 2 we set equal to 0 x

the 2 passes to subtract so here we are

dividing by x minus two let's see

what the result was the division since

the divisor polynomial was of third degree

now it will be of second degree

so it would be 1 x to the 2

+ 5 x

+ 6

notice x to the 2 x 1 and it ended

independent this expression

multiplied by x 2

gives me this expression that I had here now

x x 1 gives me the original expression we already

have 2 polynomials of first degree and 1

of second degree if you want you can

apply synthetic division with the

values ​​that we have left that are

divisors of 6 but here it is faster

if we apply the case of factorization

trinomial of the form x squared plus

bx plus c is much simpler since We

have a second-degree polynomial,

so I open the parentheses, take the square

root of the first one,

copy the plus sign, and here, plus times plus

gives me plus. I need two numbers that

multiply to give me 6 and add up to

5 because the signs are the same. The

numbers are 3 and 2. Remember to

always put the larger number in the first

parenthesis. We

bring down x2 and we bring down x1. This

means we have already managed to factor

this entire expression into four

first-degree polynomials. First, we did it as 9

third-degree and 1 first-degree, 1 second-degree and 2 first-degree, and

now we have 4 first-degree polynomials.

So, now that we have factored, we can

analyze our equation. We

already have our factored polynomial,

and on the other side, we have zero. So,

we can analyze polynomial by

polynomial. We have x + 3,

x²,

x²,

and x¹. I'm going

to draw a number line for each one.

Now I'm going to draw four

vertical lines because we have four

polynomials,

and now we are going to find the zeros of each

polynomial. This polynomial has at

least 0. 3, since negative three plus three equals

zero, this number minus two equals two

positive,

and this one equals one positive, since one

minus 10 equals ten. These numbers are the roots of

this polynomial; they are those by which,

doing synthetic division, I got

an exact division. Now I'm going to

place them as they go on the number line: to

the left, negative 3, then negative 2,

then positive 1, and to the right,

positive 2. Let's analyze the signs

of each polynomial. This polynomial is

made 0 by negative 3. Then this line will

divide its signs, as x is

positive, everything to the right is positive

and everything to its left is negative. Let's go

with x², x + 2 makes 0 in the number negative

2. Then it divides its signs into two

parts, as x is positive, everything to

the right is positive and everything to the

left must be negative. Now let's go

with x², the number 2 is what divides it,

as x is positive, everything to the

right of 2 is positive and everything to the

left is negative.

And finally, x¹, the number that makes it 0

is positive 1, and as x is

positive, everything to its right is negative. Positive and

all the left negative.

Remember that if any of these x's

were negative, we put a minus sign to the

right and a plus sign to the left.

Now, since we have multiplication, we're going

to multiply the signs, but remember

that it's enough to multiply only the minus signs, as they are the ones

that alter the result in

a multiplication of signs. The

plus sign doesn't affect the result. So,

minus x minus plus plus plus x minus minus and minus times minus

is plus, minus times minus plus and plus times minus

is minus, minus times minus plus.

Here I only have one minus sign, and

here everything is positive, so the result

is positive. Now we notice that they ask for

the numbers greater than or equal to 0;

those are the positive numbers. So,

the interval that goes from negative

infinity to negative 3 works for me,

and I make it a closed interval because

here they give me the option that it can be the

same as the union of the interval that goes from negative 2

to 1, which is also positive,

and we join that to this interval that goes

from 2 to infinity.

This would already be the solution set for

the proposed situation.

I hope you understood the topic we

tried to explain in this tutorial. If

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hope you're doing very well. Until next time. a

future video