Solution factorization of a square trinomial. Factorization of square trinomials: examples and formulas. Examples of factoring polynomials using formulas

A square trinomial is a polynomial of the form ax^2+bx+c, where x is a variable, a, b and c are some numbers, and a is not equal to zero.
Actually, the first thing we need to know in order to factorize the ill-fated trinomial is the theorem. It looks like this: “If x1 and x2 are roots square trinomial ax^2+bx+c, then ax^2+bx+c=a(x-x1)(x-x2)”. Of course, there is also a proof of this theorem, but it requires some theoretical knowledge (if we take out the factor a in the polynomial ax^2+bx+c we get ax^2+bx+c=a(x^2+(b/a) x + c/a) By Viette's theorem x1+x2=-(b/a), x1*x2=c/a, hence b/a=-(x1+x2), c/a=x1*x2. , x^2+ (b/a)x+c/a= x^2- (x1+x2)x+ x1x2=x^2-x1x-x2x+x1x2=x(x-x1)-x2(x-x1 )= (x-x1)(x-x2), so ax^2+bx+c=a(x-x1)(x-x2) Sometimes teachers make you learn the proof, but if it is not required, I advise you to just remember final formula.

2 step

Let's take as an example the trinomial 3x^2-24x+21. The first thing we need to do is equate the trinomial to zero: 3x^2-24x+21=0. The roots of the resulting quadratic equation will be the roots of the trinomial, respectively.

3 step

Solve the equation 3x^2-24x+21=0. a=3, b=-24, c=21. So, let's decide. Who doesn't know how to decide quadratic equations, look at my instruction with 2 ways to solve them using the same equation as an example. We got the roots x1=7, x2=1.

4 step

Now that we have the trinomial roots, we can safely substitute them into the formula =) ax^2+bx+c=a(x-x1)(x-x2)
we get: 3x^2-24x+21=3(x-7)(x-1)
You can get rid of the term a by putting it in brackets: 3x^2-24x+21=(x-7)(x*3-1*3)
as a result we get: 3x^2-24x+21=(x-7)(3x-3). Note: each of the obtained factors ((x-7), (3x-3) are polynomials of the first degree. That's the whole expansion =) If you doubt the answer you got, you can always check it by multiplying the brackets.

5 step

Verification of the solution. 3x^2-24x+21=3(x-7)(x-3)
(x-7)(3x-3)=3x^2-3x-21x+21=3x^2-24x+21. Now we know for sure that our solution is correct! I hope my instructions help someone =) Good luck with your studies!

In our case, in the equation D > 0 and we got 2 roots each. If it were D<0, то уравнение, как и многочлен, соответственно, корней бы не имело.
If a square trinomial has no roots, then it cannot be factored into factors that are polynomials of the first degree.

In this lesson, we will learn how to decompose square trinomials into linear factors. For this, it is necessary to recall Vieta's theorem and its inverse. This skill will help us quickly and conveniently decompose square trinomials into linear factors, and also simplify the reduction of fractions consisting of expressions.

So back to the quadratic equation , where .

What we have on the left side is called the square trinomial.

The theorem is true: If are the roots of a square trinomial, then the identity is true

Where is the leading coefficient, are the roots of the equation.

So, we have a quadratic equation - a square trinomial, where the roots of the quadratic equation are also called the roots of the quadratic trinomial. Therefore, if we have the roots of a square trinomial, then this trinomial is decomposed into linear factors.

Proof:

The proof of this fact is carried out using the Vieta theorem, which we considered in previous lessons.

Let's remember what Vieta's theorem tells us:

If are the roots of a square trinomial for which , then .

This theorem implies the following assertion that .

We see that, according to the Vieta theorem, i.e., substituting these values into the formula above, we get the following expression

Q.E.D.

Recall that we proved the theorem that if are the roots of a square trinomial, then the decomposition is valid.

Now let's recall an example of a quadratic equation, to which we selected the roots using Vieta's theorem. From this fact we can obtain the following equality thanks to the proved theorem:

Now let's check the correctness of this fact by simply expanding the brackets:

We see that we factored correctly, and any trinomial, if it has roots, can be factored into linear factors according to this theorem by the formula

However, let's check whether for any equation such a factorization is possible:

Let's take the equation for example. First, let's check the sign of the discriminant

And we remember that in order to fulfill the theorem we have learned, D must be greater than 0, therefore, in this case, factoring according to the studied theorem is impossible.

Therefore, we formulate a new theorem: if a square trinomial has no roots, then it cannot be decomposed into linear factors.

So, we have considered the Vieta theorem, the possibility of decomposing a square trinomial into linear factors, and now we will solve several problems.

Task #1

In this group, we will actually solve the problem inverse to the one posed. We had an equation, and we found its roots, decomposing into factors. Here we will do the opposite. Let's say we have the roots of a quadratic equation

The inverse problem is this: write a quadratic equation so that were its roots.

There are 2 ways to solve this problem.

Since are the roots of the equation, then is a quadratic equation whose roots are given numbers. Now let's open the brackets and check:

This was the first way we created a quadratic equation with given roots that does not have any other roots, since any quadratic equation has at most two roots.

This method involves the use of the inverse Vieta theorem.

If are the roots of the equation, then they satisfy the condition that .

For the reduced quadratic equation , , i.e. in this case , and .

Thus, we have created a quadratic equation that has the given roots.

Task #2

You need to reduce the fraction.

We have a trinomial in the numerator and a trinomial in the denominator, and the trinomials may or may not be factorized. If both the numerator and the denominator are factorized, then among them there may be equal factors that can be reduced.

First of all, it is necessary to factorize the numerator.

First, you need to check whether this equation can be factored, find the discriminant . Since , then the sign depends on the product ( must be less than 0), in this example , i.e., the given equation has roots.

To solve, we use the Vieta theorem:

In this case, since we are dealing with roots, it will be quite difficult to simply pick up the roots. But we see that the coefficients are balanced, i.e. if we assume that , and substitute this value into the equation, then the following system is obtained: i.e. 5-5=0. Thus, we have chosen one of the roots of this quadratic equation.

We will look for the second root by substituting what is already known into the system of equations, for example, , i.e. .

Thus, we have found both roots of the quadratic equation and can substitute their values into the original equation to factor it:

Recall the original problem, we needed to reduce the fraction.

Let's try to solve the problem by substituting instead of the numerator .

It is necessary not to forget that in this case the denominator cannot be equal to 0, i.e.,.

If these conditions are met, then we have reduced the original fraction to the form .

Task #3 (task with a parameter)

At what values of the parameter is the sum of the roots of the quadratic equation

If the roots of this equation exist, then , the question is when .

This online calculator is designed to factorize a function.

For example, factorize: x 2 /3-3x+12 . Let's write it as x^2/3-3*x+12 . You can also use this service, where all calculations are saved in Word format.

For example, decompose into terms. Let's write it as (1-x^2)/(x^3+x) . To see the progress of the solution, click Show steps . If you need to get the result in Word format, use this service.

Note: the number "pi" (π) is written as pi ; square root as sqrt , e.g. sqrt(3) , the tangent of tg is written as tan . See the Alternative section for a response.

If a simple expression is given, for example, 8*d+12*c*d , then factoring the expression means to factor the expression. To do this, you need to find common factors. We write this expression as: 4*d*(2+3*c) .
Express the product as two binomials: x 2 + 21yz + 7xz + 3xy . Here we already need to find several common factors: x(x + 7z) + 3y(x + 7z). We take out (x+7z) and get: (x+7z)(x + 3y) .

see also Division of polynomials by a corner (all steps of division by a column are shown)

Useful in learning the rules of factorization are abbreviated multiplication formulas, with which it will be clear how to open brackets with a square:

(a+b) 2 = (a+b)(a+b) = a 2 +2ab+b 2
(a-b) 2 = (a-b)(a-b) = a 2 -2ab+b 2
(a+b)(a-b) = a 2 - b 2
a 3 +b 3 = (a+b)(a 2 -ab+b 2)
a 3 -b 3 = (a-b)(a 2 +ab+b 2)
(a+b) 3 = (a+b)(a+b) 2 = a 3 +3a 2 b + 3ab 2 +b 3
(a-b) 3 = (a-b)(a-b) 2 = a 3 -3a 2 b + 3ab 2 -b 3

Factoring methods

After learning a few tricks factorization solutions can be classified as follows:

Using abbreviated multiplication formulas.
Search for a common factor.

This is one of the most elementary ways to simplify an expression. To apply this method, let's remember the distributive law of multiplication with respect to addition (do not be afraid of these words, you definitely know this law, you just might have forgotten its name).

The law says: in order to multiply the sum of two numbers by a third number, you need to multiply each term by this number and add the results, in other words,.

You can also do the reverse operation, and it is this reverse operation that interests us. As can be seen from the sample, the common factor a, can be taken out of the bracket.

A similar operation can be done both with variables, such as and, for example, and with numbers: .

Yes, this is too elementary an example, just like the example given earlier, with the expansion of a number, because everyone knows what numbers are, and are divisible by, but what if you got a more complicated expression:

How to find out what, for example, a number is divided into, no, with a calculator, anyone can, but without it it’s weak? And for this there are signs of divisibility, these signs are really worth knowing, they will help you quickly understand whether it is possible to take the common factor out of the bracket.

Signs of divisibility

It is not so difficult to remember them, most likely, most of them were already familiar to you, and something will be a new useful discovery, more details in the table:

Note: The table lacks a sign of divisibility by 4. If the last two digits are divisible by 4, then the whole number is divisible by 4.

Well, how do you like the sign? I advise you to remember it!

Well, let's get back to the expression, maybe take it out of the bracket and that's enough from it? No, it is customary for mathematicians to simplify, so to the fullest, take out EVERYTHING that is taken out!

And so, everything is clear with the player, but what about the numerical part of the expression? Both numbers are odd, so you can't divide by

You can use the sign of divisibility by, the sum of the digits, and, of which the number consists, is equal, and is divisible by, which means it is divisible by.

Knowing this, you can safely divide into a column, as a result of dividing by we get (signs of divisibility came in handy!). Thus, we can take the number out of the bracket, just like y, and as a result we have:

To make sure that everything is decomposed correctly, you can check the expansion by multiplication!

Also, the common factor can be taken out in power expressions. Here, for example, do you see the common factor?

All members of this expression have x's - we take out, all are divided by - we take out again, we look at what happened: .

2. Abbreviated multiplication formulas

Abbreviated multiplication formulas have already been mentioned in theory, if you can hardly remember what it is, then you should refresh them in your memory.

Well, if you consider yourself very smart and you are too lazy to read such a cloud of information, then just read on, look at the formulas and immediately take on the examples.

The essence of this decomposition is to notice some definite formula in the expression before you, apply it and thus obtain the product of something and something, that's all the decomposition. Following are the formulas:

Now try factoring the following expressions using the above formulas:

And here is what should have happened:

As you have noticed, these formulas are a very effective way of factoring, it is not always suitable, but it can be very useful!

3. Grouping or grouping method

Here's another example for you:

Well, what are you going to do with it? It seems to be divisible by and into something, and something into and into

But you can’t divide everything together into one thing, well there is no common factor, how not to look for what, and leave it without factoring?

Here you need to show ingenuity, and the name of this ingenuity is a grouping!

It is used just when not all members have common divisors. For grouping you need find groups of terms that have common divisors and rearrange them so that the same multiplier can be obtained from each group.

Of course, it is not necessary to rearrange in places, but this gives visibility, for clarity, you can take individual parts of the expression in brackets, it is not forbidden to put them as much as you like, the main thing is not to confuse the signs.

All this is not very clear? Let me explain with an example:

In a polynomial - put a member - after the member - we get

we group the first two terms together in a separate bracket and group the third and fourth terms in the same way, leaving the minus sign out of the bracket, we get:

And now we look separately at each of the two "heaps" into which we have broken the expression with brackets.

The trick is to break it into such piles from which it will be possible to take out the largest possible factor, or, as in this example, try to group the members so that after taking the factors out of the brackets from the piles, we have the same expressions inside the brackets.

From both brackets we take out the common factors of the members, from the first bracket, and from the second bracket, we get:

But it's not decomposition!

Pdonkey decomposition should remain only multiplication, but for now we have a polynomial simply divided into two parts ...

BUT! This polynomial has a common factor. This

outside the bracket and we get the final product

Bingo! As you can see, there is already a product and outside the brackets there is neither addition nor subtraction, the decomposition is completed, because we have nothing more to take out of the brackets.

It may seem like a miracle that after taking the factors out of the brackets, we still have the same expressions in the brackets, which, again, we took out of the brackets.

And this is not a miracle at all, the fact is that the examples in textbooks and in the exam are specially made in such a way that most of the expressions in tasks for simplification or factorization with the right approach to them, they are easily simplified and abruptly collapse like an umbrella when you press a button, so look for that very button in each expression.

Something I digress, what do we have there with simplification? The intricate polynomial took on a simpler form: .

Agree, not as bulky as it used to be?

4. Selection of a full square.

Sometimes, in order to apply the formulas for abbreviated multiplication (repeat the topic), it is necessary to transform the existing polynomialby presenting one of its terms as the sum or difference of two terms.

In which case you have to do this, you will learn from the example:

A polynomial in this form cannot be decomposed using abbreviated multiplication formulas, so it must be converted. Perhaps at first it will not be obvious to you which term to divide into which, but over time you will learn to immediately see the formulas for abbreviated multiplication, even if they are not present in their entirety, and you will quickly determine what is missing here to the full formula, but for now - learn , a student, more precisely a schoolboy.

For the full formula of the square of the difference, here you need instead. Let's represent the third term as a difference, we get: We can apply the difference square formula to the expression in brackets (not to be confused with the difference of squares!!!), we have: , to this expression, we can apply the formula for the difference of squares (not to be confused with the squared difference!!!), imagining how, we get: .

An expression not always factored into factors looks simpler and smaller than it was before decomposition, but in this form it becomes more mobile, in the sense that you can not worry about changing signs and other mathematical nonsense. Well, for you to decide on your own, the following expressions need to be factored.

Examples:

Answers:

5. Factorization of a square trinomial

For the factorization of a square trinomial, see below in the decomposition examples.

Examples of 5 Methods for Factoring a Polynomial

1. Taking the common factor out of brackets. Examples.

Do you remember what the distributive law is? This is such a rule:

Example:

Factorize a polynomial.

Solution:

Another example:

Multiply.

Solution:

If the whole term is taken out of brackets, one remains in brackets instead of it!

2. Formulas for abbreviated multiplication. Examples.

The most commonly used formulas are the difference of squares, the difference of cubes and the sum of cubes. Remember these formulas? If not, urgently repeat the topic!

Example:

Factor the expression.

Solution:

In this expression, it is easy to find out the difference of cubes:

Example:

Solution:

3. Grouping method. Examples

Sometimes it is possible to interchange the terms in such a way that one and the same factor can be extracted from each pair of neighboring terms. This common factor can be taken out of the bracket and the original polynomial will turn into a product.

Example:

Factor out the polynomial.

Solution:

We group the terms as follows:
.

In the first group, we take the common factor out of brackets, and in the second - :
.

Now the common factor can also be taken out of brackets:
.

4. The method of selection of a full square. Examples.

If the polynomial can be represented as the difference of the squares of two expressions, all that remains is to apply the abbreviated multiplication formula (difference of squares).

Example:

Factor out the polynomial.

Solution:Example:

\begin(array)(*(35)(l))
((x)^(2))+6(x)-7=\underbrace(((x)^(2))+2\cdot 3\cdot x+9)_(square\ sums\ ((\left (x+3 \right))^(2)))-9-7=((\left(x+3 \right))^(2))-16= \\
=\left(x+3+4 \right)\left(x+3-4 \right)=\left(x+7 \right)\left(x-1 \right) \\
\end(array)

Factor out the polynomial.

Solution:

\begin(array)(*(35)(l))
((x)^(4))-4((x)^(2))-1=\underbrace(((x)^(4))-2\cdot 2\cdot ((x)^(2) )+4)_(square\ differences((\left(((x)^(2))-2 \right))^(2)))-4-1=((\left(((x)^ (2))-2 \right))^(2))-5= \\
=\left(((x)^(2))-2+\sqrt(5) \right)\left(((x)^(2))-2-\sqrt(5) \right) \\
\end(array)

5. Factorization of a square trinomial. Example.

A square trinomial is a polynomial of the form, where is an unknown, are some numbers, moreover.

Variable values that turn the square trinomial to zero are called roots of the trinomial. Therefore, the roots of a trinomial are the roots of a quadratic equation.

Theorem.

Example:

Let's factorize the square trinomial: .

First, we solve the quadratic equation: Now we can write the factorization of this square trinomial into factors:

Now your opinion...

We have described in detail how and why to factorize a polynomial.

We gave a lot of examples of how to do it in practice, pointed out the pitfalls, gave solutions ...

What do you say?

How do you like this article? Do you use these tricks? Do you understand their essence?

Write in the comments and... get ready for the exam!

So far, it's the most important thing in your life.