Lesson “Application of various methods for factoring a polynomial. Application of various methods for factoring a polynomial Application of various methods for factoring a polynomial

LESSON PLAN algebra lesson in 7th grade

Teacher Prilepova O.A.

Lesson Objectives:

Show the application of various methods for factoring a polynomial

Repeat the methods of factorization and consolidate their knowledge during the exercises

To develop the skills and abilities of students in the application of abbreviated multiplication formulas.

Develop students' logical thinking and interest in the subject.

Tasks:

in the direction personal development:

Development of interest in mathematical creativity and mathematical abilities;

Development of initiative, activity in solving mathematical problems;

Cultivating the ability to make independent decisions.

in the meta-subject direction :

Formation of general ways of intellectual activity, characteristic of mathematics and being the basis of cognitive culture;

Use of ICT technology;

in the subject area:

Mastering the mathematical knowledge and skills necessary to continue education;

Formation in students the ability to look for ways to factorize a polynomial and find them for a polynomial that is factorized.

Equipment:handouts, route sheets with evaluation criteria,multimedia projector, presentation.

Lesson type:repetition, generalization and systematization of the material covered

Forms of work:work in pairs and groups, individual, collective,independent, frontal work.

During the classes:

Exists several different ways factorization of a polynomial. Most often, in practice, not one, but several methods are used at once. There can be no specific order of actions here, in each example everything is individual. But you can try to follow the following order:

1. If there is a common factor, then take it out of the bracket;

2. After that, try to factorize the polynomial using the abbreviated multiplication formulas;

3. If after that we have not yet received the desired result, we should try to use the grouping method.

Abbreviated multiplication formulas

1. a^2 - b^2 = (a+b)*(a-b);

2. (a+b)^2 = a^2+2*a*b+b^2;

3. (a-b)^2 = a^2-2*a*b+b^2;

4. a^3+b^3 = (a+b)*(a^2 - a*b+b^2);

5. a^3 - b^3 = (a-b)*(a^2 + a*b+b^2);

Now let's take a look at a few examples:

Example 1

Factorize the polynomial: (a^2+1)^2 - 4*a^2

First, we apply the abbreviated multiplication formula "difference of squares" and open the inner brackets.

(a^2+1)^2 - 4*a^2 = ((a^2+1)-2*a)*((a^2+1)+2*a) = (a^2+1 -2*a)*(a^2+1+2*a);

Note that the expressions for the square of the sum and the square of the difference of two expressions are obtained in brackets. Apply them and get the answer.

a^2+1-2*a)*(a^2+1+2*a) = (a-1)^2*(a+1)^2;

Answer:(a-1)^2*(a+1)^2;

Example 2

Factorize the polynomial 4*x^2 - y^2 + 4*x +2*y.

As you can see directly here, none of the methods is suitable. But there are two squares, they can be grouped. Let's try.

4*x^2 - y^2 + 4*x +2*y = (4*x^2 - y^2) +(4*x +2*y);

We got the formula for the difference of squares in the first bracket, And in the second bracket there is a common factor of two. Let's apply the formula and take out the common factor.

(4*x^2 - y^2) +(4*x +2*y)= (2*x - y)*(2*x+y) +2*(2*x+y);

It can be seen that two identical brackets are obtained. We take them out as a common factor.

(2*x - y)*(2*x+y) +2*(2*x+y) = (2*x+y)*(2*x - y)+2)= (2*x+ y)*(2*x-y+2);

Answer:(2*x+y)*(2*x-y+2);

As you can see, there is no universal way. With experience, the skill will come and factoring the polynomial into factors will be very easy.

In the previous lesson, we studied the multiplication of a polynomial by a monomial. For example, the product of a monomial a and a polynomial b + c is found like this:

a(b + c) = ab + bc

However, in some cases it is more convenient to perform the inverse operation, which can be called taking the common factor out of brackets:

ab + bc = a(b + c)

For example, suppose we need to calculate the value of the polynomial ab + bc with the values ​​of the variables a = 15.6, b = 7.2, c = 2.8. If we substitute them directly into the expression, we get

ab + bc = 15.6 * 7.2 + 15.6 * 2.8

ab + bc = a(b + c) = 15.6 * (7.2 + 2.8) = 15.6 * 10 = 156

In this case, we have represented the polynomial ab + bc as the product of two factors: a and b + c. This action is called the factorization of a polynomial.

Moreover, each of the factors into which the polynomial is decomposed can, in turn, be a polynomial or a monomial.

Consider the polynomial 14ab - 63b 2 . Each of its constituent monomials can be represented as a product:

It can be seen that both polynomials have a common factor 7b. So, it can be taken out of brackets:

14ab - 63b 2 = 7b*2a - 7b*9b = 7b(2a-9b)

You can check the correctness of taking the factor out of the brackets using the inverse operation - expanding the bracket:

7b(2a - 9b) = 7b*2a - 7b*9b = 14ab - 63b 2

It is important to understand that often a polynomial can be expanded in several ways, for example:

5abc + 6bcd = b(5ac + 6cd) = c(5ab + 6bd) = bc(5a + 6d)

Usually they try to endure, roughly speaking, the "greatest" monomial. That is, the polynomial is laid out in such a way that nothing more can be taken out of the remaining polynomial. So, when splitting

5abc + 6bcd = b(5ac + 6cd)

the sum of monomials that have a common factor c remains in brackets. If we also take it out, then there will be no common factors in brackets:

b(5ac + 6cd) = bc(5a + 6d)

Let us analyze in more detail how to find common factors for monomials. Let's split the sum

8a 3 b 4 + 12a 2 b 5 v + 16a 4 b 3 c 10

It consists of three components. First, let's look at the numerical coefficients in front of them. These are 8, 12 and 16. In lesson 3 of grade 6, the topic of GCD and the algorithm for finding it were considered. This is the greatest common divisor. You can almost always pick it up orally. The numerical coefficient of the common factor will just be the GCD of the numerical coefficients of the terms of the polynomial. In this case, the number is 4.

Next, we look at the degrees of these variables. In the common factor, the letters must have the minimum degrees that occur in terms. So, the variable a in a polynomial of degree 3, 2, and 4 (minimum 2), so the common factor will be a 2 . The variable b has a minimum degree of 3, so the common factor will be b 3:

8a 3 b 4 + 12a 2 b 5 v + 16a 4 b 3 c 10 = 4a 2 b 3 (2ab + 3b 2 c + 4a 2 c 10)

As a result, the remaining terms 2ab, 3b 2 c, 4a 2 c 10 have no common letter variable, and their coefficients 2, 3, and 4 have no common divisors.

You can take out of brackets not only monomials, but also polynomials. For example:

x(a-5) + 2y(a-5) = (a-5)(x+2y)

One more example. It is necessary to expand the expression

5t(8y - 3x) + 2s(3x - 8y)

Solution. Recall that the minus sign reverses the signs in brackets, so

-(8y - 3x) = -8y + 3x = 3x - 8y

So you can replace (3x - 8y) with - (8y - 3x):

5t(8y - 3x) + 2s(3x - 8y) = 5t(8y - 3x) + 2*(-1)s(8y - 3x) = (8y - 3x)(5t - 2s)

Answer: (8y - 3x)(5t - 2s).

Remember that the subtracted and reduced can be interchanged by changing the sign in front of the brackets:

(a - b) = - (b - a)

The opposite is also true: the minus already in front of the brackets can be removed if the subtracted and reduced are rearranged at the same time:

This technique is often used in problem solving.

Grouping method

Consider another way to factorize a polynomial, which helps to factorize a polynomial. Let there be an expression

ab - 5a + bc - 5c

It is not possible to take out a factor that is common to all four monomials. However, you can represent this polynomial as the sum of two polynomials, and in each of them take the variable out of brackets:

ab - 5a + bc - 5c = (ab - 5a) + (bc - 5c) = a(b - 5) + c(b - 5)

Now you can take out the expression b - 5:

a(b - 5) + c(b - 5) = (b - 5)(a + c)

We "grouped" the first term with the second, and the third with the fourth. Therefore, the described method is called the grouping method.

Example. Let us expand the polynomial 6xy + ab- 2bx- 3ay.

Solution. Grouping the 1st and 2nd terms is impossible, since they do not have a common factor. So let's swap the monomials:

6xy + ab - 2bx - 3ay = 6xy - 2bx + ab - 3ay = (6xy - 2bx) + (ab - 3ay) = 2x(3y - b) + a(b - 3y)

The differences 3y - b and b - 3y differ only in the order of the variables. In one of the brackets, it can be changed by moving the minus sign out of the brackets:

(b - 3y) = - (3y - b)

We use this substitution:

2x(3y - b) + a(b - 3y) = 2x(3y - b) - a(3y - b) = (3y - b)(2x - a)

The result is an identity:

6xy + ab - 2bx - 3ay = (3y - b)(2x - a)

Answer: (3y - b)(2x - a)

You can group not only two, but in general any number of terms. For example, in the polynomial

x 2 - 3xy + xz + 2x - 6y + 2z

you can group the first three and last 3 monomials:

x 2 - 3xy + xz + 2x - 6y + 2z = (x 2 - 3xy + xz) + (2x - 6y + 2z) = x(x - 3y + z) + 2(x - 3y + z) = (x + 2)(x - 3y + z)

Now let's look at the task of increased complexity

Example. Expand the square trinomial x 2 - 8x +15.

Solution. This polynomial consists of only 3 monomials, and therefore, as it seems, the grouping cannot be done. However, you can make the following substitution:

Then the original trinomial can be represented as follows:

x 2 - 8x + 15 = x 2 - 3x - 5x + 15

Let's group the terms:

x 2 - 3x - 5x + 15 = (x 2 - 3x) + (- 5x + 15) = x(x - 3) - 5(x - 3) = (x - 5)(x - 3)

Answer: (x - 5) (x - 3).

Of course, guessing about the replacement - 8x = - 3x - 5x in the above example is not easy. Let's show a different line of reasoning. We need to expand the polynomial of the second degree. As we remember, when multiplying polynomials, their degrees are added. This means that if we can decompose the square trinomial into two factors, then they will be two polynomials of the 1st degree. Let's write the product of two polynomials of the first degree, whose leading coefficients are equal to 1:

(x + a)(x + b) = x 2 + xa + xb + ab = x 2 + (a + b)x + ab

Here a and b are some arbitrary numbers. In order for this product to be equal to the original trinomial x 2 - 8x +15, it is necessary to choose the appropriate coefficients for the variables:

With the help of selection, it can be determined that the numbers a= - 3 and b = - 5 satisfy this condition. Then

(x - 3)(x - 5) = x 2 * 8x + 15

which can be verified by opening the brackets.

For simplicity, we considered only the case when the multiplied polynomials of the 1st degree have the highest coefficients equal to 1. However, they could be equal, for example, to 0.5 and 2. In this case, the decomposition would look somewhat different:

x 2 * 8x + 15 = (2x - 6)(0.5x - 2.5)

However, by taking the factor 2 out of the first bracket and multiplying it by the second, we would get the original expansion:

(2x - 6)(0.5x - 2.5) = (x - 3) * 2 * (0.5x - 2.5) = (x - 3)(x - 5)

In the considered example, we decomposed the square trinomial into two polynomials of the first degree. In the future, we will often have to do this. However, it is worth noting that some square trinomials, for example,

it is impossible to decompose in this way into a product of polynomials. This will be proven later.

Application of factorization of polynomials

Factoring a polynomial can simplify some operations. Let it be necessary to evaluate the value of the expression

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8 + 2 9

We take out the number 2, while the degree of each term decreases by one:

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8 + 2 9 = 2(1 + 2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8)

Denote the sum

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8

for x. Then the above equation can be rewritten:

x + 2 9 = 2(1 + x)

We got the equation, we will solve it (see the lesson of the equation):

x + 2 9 = 2(1 + x)

x + 2 9 = 2 + 2x

2x - x = 2 9 - 2

x = 512 - 2 = 510

Now let's express the amount we are looking for in terms of x:

2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8 + 2 9 = x + 2 9 = 510 + 512 = 1022

When solving this problem, we raised the number 2 only to the 9th power, and we managed to exclude all other exponentiation operations from calculations by factoring the polynomial. Similarly, you can make a calculation formula for other similar amounts.

Now let's calculate the value of the expression

38.4 2 - 61.6 * 29.5 + 61.6 * 38.4 - 29.5 * 38.4

38.4 2 - 61.6 * 29.5 + 61.6 * 38.4 - 29.5 * 38.4 = 38.4 2 - 29.5 * 38.4 + 61.6 * 38.4 - 61.6 * 29.5 = 38.4(38.4 - 29.5) + 61.6(38.4 - 29.5) = (38.4 + 61.6)(38.4 - 29.5) = 8.9*100 = 890

81 4 - 9 7 + 3 12

is divisible by 73. Note that the numbers 9 and 81 are powers of three:

81 = 9 2 = (3 2) 2 = 3 4

Knowing this, we will make a replacement in the original expression:

81 4 - 9 7 + 3 12 = (3 4) 4 - (3 2) 7 + 3 12 = 3 16 - 3 14 + 3 12

Let's take out 3 12:

3 16 - 3 14 + 3 12 = 3 12 (3 4 - 3 2 + 1) = 3 12 * (81 - 9 + 1) = 3 12 * 73

The product 3 12 .73 is divisible by 73 (since one of the factors is divisible by it), so the expression 81 4 - 9 7 + 3 12 is divisible by this number.

Factoring out can be used to prove identities. For example, let us prove the validity of the equality

(a 2 + 3a) 2 + 2(a 2 + 3a) = a(a + 1)(a + 2)(a + 3)

To solve the identity, we transform the left side of the equality by taking out the common factor:

(a 2 + 3a) 2 + 2(a 2 + 3a) = (a 2 + 3a)(a 2 + 3a) + 2(a 2 + 3a) = (a 2 + 3a)(a 2 + 3a + 2 )

(a 2 + 3a)(a 2 + 3a + 2) = (a 2 + 3a)(a 2 + 2a + a + 2) = (a 2 + 3a)((a 2 + 2a) + (a + 2 ) = (a 2 + 3a)(a(a + 2) + (a + 2)) = (a 2 + 3a)(a + 1)(a + 2) = a(a + 3)(a + z )(a + 2) = a(a + 1)(a + 2)(a + 3)

One more example. Let us prove that for any values ​​of the variables x and y, the expression

(x - y)(x + y) - 2x(x - y)

is not a positive number.

Solution. Let's take out the common factor x - y:

(x - y)(x + y) - 2x(x - y) = (x - y)(x + y - 2x) = (x - y)(y - x)

Note that we have obtained the product of two similar binomials that differ only in the order of the letters x and y. If we swapped the variables in one of the brackets, we would get the product of two identical expressions, that is, a square. But in order to swap x and y, you need to put a minus sign in front of the bracket:

(x - y) = -(y - x)

Then you can write:

(x - y)(y - x) = -(y - x)(y - x) = -(y - x) 2

As you know, the square of any number is greater than or equal to zero. This also applies to the expression (y - x) 2 . If there is a minus before the expression, then it must be less than or equal to zero, that is, it is not a positive number.

Polynomial expansion helps solve some equations. This uses the following statement:

If in one part of the equation there is zero, and in the other the product of factors, then each of them should be equated to zero.

Example. Solve the equation (s - 1)(s + 1) = 0.

Solution. The product of monomials s - 1 and s + 1 is written on the left side, and zero is written on the right. Therefore, either s - 1 or s + 1 must equal zero:

(s - 1)(s + 1) = 0

s - 1 = 0 or s + 1 = 0

s=1 or s=-1

Each of the two obtained values ​​of the variable s is the root of the equation, that is, it has two roots.

Answer: -1; one.

Example. Solve the equation 5w 2 - 15w = 0.

Solution. Let's take out 5w:

Again, the product is written on the left side, and zero on the right. Let's continue with the solution:

5w = 0 or (w - 3) = 0

w=0 or w=3

Answer: 0; 3.

Example. Find the roots of the equation k 3 - 8k 2 + 3k- 24 = 0.

Solution. Let's group the terms:

k 3 - 8k 2 + 3k- 24 = 0

(k 3 - 8k 2) + (3k - 24) = 0

k 2 (k - 8) + 3(k - 8) = 0

(k 3 + 3)(k - 8) = 0

k 2 + 3 = 0 or k - 8 = 0

k 2 \u003d -3 or k \u003d 8

Note that the equation k 2 = - 3 has no solution, since any number squared is not less than zero. Therefore, the only root of the original equation is k = 8.

Example. Find the roots of the equation

(2u - 5)(u + 3) = 7u + 21

Solution: Move all the terms to the left side, and then group the terms:

(2u - 5)(u + 3) = 7u + 21

(2u - 5)(u + 3) - 7u - 21 = 0

(2u - 5)(u + 3) - 7(u + 3) = 0

(2u - 5 - 7)(u + 3) = 0

(2u - 12)(u + 3) = 0

2u - 12 = 0 or u + 3 = 0

u=6 or u=-3

Answer: - 3; 6.

Example. Solve the Equation

(t 2 - 5t) 2 = 30t - 6t 2

(t 2 - 5t) 2 = 30t - 6t 2

(t 2 - 5t) 2 - (30t - 6t 2) = 0

(t 2 - 5t)(t 2 - 5t) + 6(t 2 - 5t) = 0

(t 2 - 5t)(t 2 - 5t + 6) = 0

t 2 - 5t = 0 or t 2 - 5t + 6 = 0

t = 0 or t - 5 = 0

t=0 or t=5

Now let's take a look at the second equation. Before us is again a square trinomial. To factorize it by the grouping method, you need to represent it as a sum of 4 terms. If we make the replacement - 5t = - 2t - 3t, then we can further group the terms:

t 2 - 5t + 6 = 0

t 2 - 2t - 3t + 6 = 0

t(t - 2) - 3(t - 2) = 0

(t - 3)(t - 2) = 0

T - 3 = 0 or t - 2 = 0

t=3 or t=2

As a result, we found that the original equation has 4 roots.

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 must 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 decomposition 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 the common factor can be bracketed.

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 expansion 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 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. it

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 simplifying 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, to apply the formulas for abbreviated multiplication (repeat the topic), it is necessary to transform the existing polynomial, 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 abbreviated multiplication formulas, 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: .

The expression not always factorized 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!

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To factorize polynomials, we used parentheses, grouping, and abbreviated multiplication formulas. Sometimes it is possible to factorize a polynomial by applying successively several methods. In this case, the transformation should, if possible, be started by taking the common factor out of brackets.

Example 1 Let us factorize the polynomial 10a 3 - 40a.

Solution: The terms of this polynomial have a common factor of 10a. Let's take this factor out of brackets:

10a 3 - 40a \u003d 10a (a 2 - 4).

The factorization can be continued by applying the difference of squares formula to the expression a 2 - 4. As a result, we obtain polynomials of lower degrees as factors.

10a (a 2 - 4) \u003d 10a (a + 2) (a - 2).

10a 3 - 40a \u003d 10a (a + 2) (a - 2).

Example 2 Factoring the polynomial

ab 3 - 3b 3 + ab 2 y - Zb 2 y.

Solution: First, we take the common factor b2 out of brackets:

ab 3 - 3b 3 + ab 2 y - 3b 2 y = b 2 (ab - 3b + ay - 3y).

Let us now try to factorize the polynomial

ab - 3b + ay - 3y.

Grouping the first term with the second and the third with the fourth, we have

ab - 3b + ay - Zu \u003d b (a - 3) + y (a - 3) \u003d (a - 3) (b + y).

Finally we get

ab 3 - Zb 3 + ab 2 y - Zb 2 y \u003d b 2 (a - 3) (b + y).

Example 3 Let us factorize the polynomial a 2 - 4ax - 9 + 4x 2.

Solution: Let's group the first, second and fourth terms of the polynomial. We get the trinomial a 2 - 4ax + 4x 2, which can be represented as a square of the difference. That's why

a 2 - 4ax - 9 + 4x 2 \u003d (a 2 - 4ax + 4x 2) - 9 \u003d (a - 2x) 2 - 9.

The resulting expression can be factored using the difference of squares formula:

(a - 2x) 2 - 9 \u003d (a - 2x) 2 - Z 2 \u003d (a - 2x - 3) (a - 2x + 3).

Consequently,

a 2 - 4ax - 9 + 4x 2 \u003d (a - 2x - 3) (a - 2x + 3).

Note that when factoring a polynomial, we mean its representation as a product of several polynomials, in which at least two factors are polynomials of non-zero degree (that is, they are not numbers).

Not every polynomial can be factorized. For example, it is impossible to factorize the polynomials x 2 + 1, 4x 2 - 2x + 1, etc.

Let's look at an example of using factorization to simplify calculations using a calculator.

Example 4 Using the calculator, we will find the value of the polynomial bx 3 + 2x 2 - 7x + 4 for x = 1.2.

Solution: If you perform the actions in the accepted order, then you first have to find the values ​​​​of the expressions x 3 5, x 2 2 and 7x, write the results on paper or enter them into the memory of the calculator, and then proceed to addition and subtraction. However, the desired result can be obtained much easier if the given polynomial is transformed as follows:

bx 3 + 2x 2 - 7x + 4 \u003d (5x 2 + 2x - 7)x + 4 \u003d ((5x + 2)x - 7)x + 4.

After doing the calculations for x = 1.2, we find that the value of the polynomial is 7.12.

Exercises

Control questions and tasks

1. Give an example of an integer expression and an expression that is not an integer.
2. What actions must be performed and in what order to represent the whole expression 4x (3 - x) 2 + (x 2 - 4) (x + 4) as a polynomial?
3. What methods of factoring polynomials do you know?