Expanding polynomials to obtain a product can sometimes seem confusing. But it's not that difficult if you understand the process step by step. The article describes in detail how to factor a quadratic trinomial.
Many people do not understand how to factor a square trinomial, and why this is done. At first it may seem like a futile exercise. But in mathematics nothing is done for nothing. The transformation is necessary to simplify the expression and ease of calculation.
A polynomial of the form – ax²+bx+c, called a quadratic trinomial. The term "a" must be negative or positive. In practice, this expression is called a quadratic equation. Therefore, sometimes they say it differently: how to expand a quadratic equation.
Interesting! A polynomial is called a square because of its largest degree, the square. And a trinomial - because of the 3 components.
Some other types of polynomials:
- linear binomial (6x+8);
- cubic quadrinomial (x³+4x²-2x+9).
Factoring a quadratic trinomial
First, the expression is equal to zero, then you need to find the values of the roots x1 and x2. There may be no roots, there may be one or two roots. The presence of roots is determined by the discriminant. You need to know its formula by heart: D=b²-4ac.
If the result D is negative, there are no roots. If positive, there are two roots. If the result is zero, the root is one. The roots are also calculated using the formula.
If, when calculating the discriminant, the result is zero, you can use any of the formulas. In practice, the formula is simply shortened: -b / 2a.
The formulas for different discriminant values are different.
If D is positive:
If D is zero:
Online calculators
There is an online calculator on the Internet. It can be used to perform factorization. Some resources provide the opportunity to view the solution step by step. Such services help to better understand the topic, but you need to try to understand it well.
Useful video: Factoring a quadratic trinomial
Examples
We suggest looking at simple examples of how to factor a quadratic equation.
Example 1
This clearly shows that the result is two x's because D is positive. They need to be substituted into the formula. If the roots turn out to be negative, the sign in the formula changes to the opposite.
We know the formula for factoring a quadratic trinomial: a(x-x1)(x-x2). We put the values in brackets: (x+3)(x+2/3). There is no number before a term in a power. This means that there is one there, it goes down.
Example 2
This example clearly shows how to solve an equation that has one root.
We substitute the resulting value:
Example 3
Given: 5x²+3x+7
First, let's calculate the discriminant, as in previous cases.
D=9-4*5*7=9-140= -131.
The discriminant is negative, which means there are no roots.
After receiving the result, you should open the brackets and check the result. The original trinomial should appear.
Alternative solution
Some people were never able to make friends with the discriminator. There is another way to factorize a quadratic trinomial. For convenience, the method is shown with an example.
Given: x²+3x-10
We know that we should get 2 brackets: (_)(_). When the expression looks like this: x²+bx+c, at the beginning of each bracket we put x: (x_)(x_). The remaining two numbers are the product that gives “c”, i.e. in this case -10. The only way to find out what numbers these are is by selection. The substituted numbers must correspond to the remaining term.
For example, multiplying the following numbers gives -10:
- -1, 10;
- -10, 1;
- -5, 2;
- -2, 5.
- (x-1)(x+10) = x2+10x-x-10 = x2+9x-10. No.
- (x-10)(x+1) = x2+x-10x-10 = x2-9x-10. No.
- (x-5)(x+2) = x2+2x-5x-10 = x2-3x-10. No.
- (x-2)(x+5) = x2+5x-2x-10 = x2+3x-10. Fits.
This means that the transformation of the expression x2+3x-10 looks like this: (x-2)(x+5).
Important! You should be careful not to confuse the signs.
Expansion of a complex trinomial
If “a” is greater than one, difficulties begin. But everything is not as difficult as it seems.
To factorize, you first need to see if anything can be factored out.
For example, given the expression: 3x²+9x-30. Here the number 3 is taken out of brackets:
3(x²+3x-10). The result is the already well-known trinomial. The answer looks like this: 3(x-2)(x+5)
How to decompose if the term that is in the square is negative? In this case, the number -1 is taken out of brackets. For example: -x²-10x-8. The expression will then look like this:
The scheme differs little from the previous one. There are just a few new things. Let's say the expression is given: 2x²+7x+3. The answer is also written in 2 brackets that need to be filled in (_)(_). In the 2nd bracket is written x, and in the 1st what is left. It looks like this: (2x_)(x_). Otherwise, the previous scheme is repeated.
The number 3 is given by the numbers:
- -1, -3;
- -3, -1;
- 3, 1;
- 1, 3.
We solve equations by substituting these numbers. The last option is suitable. This means that the transformation of the expression 2x²+7x+3 looks like this: (2x+1)(x+3).
Other cases
It is not always possible to convert an expression. With the second method, solving the equation is not required. But the possibility of transforming terms into a product is checked only through the discriminant.
It is worth practicing solving quadratic equations so that when using the formulas there are no difficulties.
Useful video: factoring a trinomial
Conclusion
You can use it in any way. But it’s better to practice both until they become automatic. Also, learning how to solve quadratic equations well and factor polynomials is necessary for those who are planning to connect their lives with mathematics. All the following mathematical topics are built on this.
The study of many physical and geometric patterns often leads to solving problems with parameters. Some universities also include equations, inequalities and their systems in exam papers, which are often very complex and require a non-standard approach to solution. At school, this one of the most difficult sections of the school algebra course is considered only in a few elective or subject courses.
In my opinion, the functional graphical method is a convenient and fast way to solve equations with a parameter.
As is known, in relation to equations with parameters there are two formulations of the problem.
- Solve the equation (for each parameter value, find all solutions to the equation).
- Find all values of the parameter for each of which the solutions to the equation satisfy the given conditions.
In this paper, a problem of the second type is considered and studied in relation to the roots of a square trinomial, the finding of which is reduced to solving a quadratic equation.
The author hopes that this work will help teachers when developing lessons and preparing students for the Unified State Exam.
1. What is a parameter
Expression of the form ah 2 + bx + c in the school algebra course they call the quadratic trinomial with respect to X, Where a, b, c are given real numbers, and, a=/= 0. The values of the variable x at which the expression becomes zero are called the roots of the square trinomial. To find the roots of a quadratic trinomial, you need to solve the quadratic equation ah 2 + bх + c = 0.
Let's remember the basic equations from the school algebra course ax + b = 0;
aх2 + bх + c = 0. When searching for their roots, the values of the variables a, b, c, included in the equation are considered fixed and given. The variables themselves are called parameters. Since there is no definition of the parameter in school textbooks, I propose to take the following simplest version as a basis.
Definition.A parameter is an independent variable, the value of which in the problem is considered to be a given fixed or arbitrary real number, or a number belonging to a predetermined set.
2. Basic types and methods for solving problems with parameters
Among tasks with parameters, the following main types of tasks can be distinguished.
- Equations that must be solved either for any value of a parameter(s) or for parameter values belonging to a pre-specified set. For example. Solve equations: ax = 1, (a – 2)x = a 2 – 4.
- Equations for which it is necessary to determine the number of solutions depending on the value of the parameter (parameters). For example. At what parameter values a the equation 4X 2 – 4ax + 1 = 0 has a single root?
- Equations for which, for the required parameter values, the set of solutions satisfies the specified conditions in the domain of definition.
For example, find the parameter values at which the roots of the equation ( a – 2)X 2
–
2ax + a + 3 =
0
positive.
The main ways to solve problems with a parameter: analytical and graphical.
Analytical- This is a method of the so-called direct solution, repeating standard procedures for finding the answer in problems without a parameter. Let's look at an example of such a task.
Task No. 1
At what values of the parameter a does the equation X 2 – 2ax + a 2 – 1 = 0 has two different roots belonging to the interval (1; 5)?
Solution
X 2 –
2ax + a 2 –
1 = 0.
According to the conditions of the problem, the equation must have two different roots, and this is only possible under the condition: D > 0.
We have: D = 4 a 2 – 2(A 2 – 1) = 4. As we can see, the discriminant does not depend on a, therefore, the equation has two different roots for any values of the parameter a. Let's find the roots of the equation: X 1 = A + 1, X 2
= A – 1
The roots of the equation must belong to the interval (1; 5), i.e.
So, at 2<A < 4 данное уравнение имеет
два различных корня, принадлежащих промежутку (1;
5)
Answer: 2<A < 4.
This approach to solving problems of the type under consideration is possible and rational in cases where the discriminant of the quadratic equation is “good”, i.e. is the exact square of any number or expression, or the roots of the equation can be found using the inverse theorem of Vieta. Then, the roots do not represent irrational expressions. Otherwise, solving problems of this type involves quite complex procedures from a technical point of view. And solving irrational inequalities requires new knowledge from the student.
Graphic- this is a method in which graphs are used in the coordinate plane (x; y) or (x; a). The clarity and beauty of this method of solution helps to find a quick way to solve the problem. Let's solve problem No. 1 graphically.
As you know from an algebra course, the roots of a quadratic equation (quadratic trinomial) are the zeros of the corresponding quadratic function: Y = X 2
– 2Oh + A 2 – 1. The graph of the function is a parabola, the branches are directed upward (the first coefficient is 1). A geometric model that meets all the requirements of the problem looks like this.
Now all that remains is to “fix” the parabola in the desired position using the necessary conditions.
- Since a parabola has two points of intersection with the axis X, then D > 0.
- The vertex of the parabola is between the vertical lines X= 1 and X= 5, therefore the abscissa of the vertex of the parabola x o belongs to the interval (1; 5), i.e.
1 <X O< 5. - We notice that at(1) > 0, at(5) > 0.
So, moving from the geometric model of the problem to the analytical one, we obtain a system of inequalities.
Answer: 2<A < 4.
As can be seen from the example, a graphical method for solving problems of the type under consideration is possible in the case when the roots are “bad”, i.e. contain a parameter under the radical sign (in this case, the discriminant of the equation is not a perfect square).
In the second solution method, we worked with the coefficients of the equation and the range of the function at = X 2 – 2Oh + A 2
– 1.
This method of solution cannot be called only graphical, because here we have to solve a system of inequalities. Rather, this method is combined: functional and graphic. Of these two methods, the latter is not only elegant, but also the most important, since it shows the relationship between all types of mathematical models: a verbal description of the problem, a geometric model - a graph of a quadratic trinomial, an analytical model - a description of a geometric model by a system of inequalities.
So, we have considered a problem in which the roots of a quadratic trinomial satisfy given conditions in the domain of definition for the desired parameter values.
What other possible conditions can the roots of a quadratic trinomial satisfy for the desired parameter values?
Consider the quadratic equation:
(1)
.
Roots of a quadratic equation(1) are determined by the formulas:
;
.
These formulas can be combined like this:
.
When the roots of a quadratic equation are known, then a polynomial of the second degree can be represented as a product of factors (factored):
.
Next we assume that are real numbers.
Let's consider discriminant of a quadratic equation:
.
If the discriminant is positive, then the quadratic equation (1) has two different real roots:
;
.
Then the factorization of the quadratic trinomial has the form:
.
If the discriminant is equal to zero, then the quadratic equation (1) has two multiple (equal) real roots:
.
Factorization:
.
If the discriminant is negative, then the quadratic equation (1) has two complex conjugate roots:
;
.
Here is the imaginary unit, ;
and are the real and imaginary parts of the roots:
;
.
Then
.
Graphic interpretation
If you plot the function
,
which is a parabola, then the points of intersection of the graph with the axis will be the roots of the equation
.
At , the graph intersects the x-axis (axis) at two points.
When , the graph touches the x-axis at one point.
When , the graph does not cross the x-axis.
Below are examples of such graphs.
Useful formulas related to quadratic equation
(f.1) ;
(f.2) ;
(f.3) .
Derivation of the formula for the roots of a quadratic equation
We carry out transformations and apply formulas (f.1) and (f.3):
,
Where
;
.
So, we got the formula for a polynomial of the second degree in the form:
.
This shows that the equation
performed at
And .
That is, and are the roots of the quadratic equation
.
Examples of determining the roots of a quadratic equation
Example 1
(1.1)
.
Solution
.
Comparing with our equation (1.1), we find the values of the coefficients:
.
We find the discriminant:
.
Since the discriminant is positive, the equation has two real roots:
;
;
.
From here we obtain the factorization of the quadratic trinomial:
.
Graph of the function y = 2 x 2 + 7 x + 3 intersects the x-axis at two points.
Let's plot the function
.
The graph of this function is a parabola. It crosses the abscissa axis (axis) at two points:
And .
These points are the roots of the original equation (1.1).
Answer
;
;
.
Example 2
Find the roots of a quadratic equation:
(2.1)
.
Solution
Let's write the quadratic equation in general form:
.
Comparing with the original equation (2.1), we find the values of the coefficients:
.
We find the discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) roots:
;
.
Then the factorization of the trinomial has the form:
.
Graph of the function y = x 2 - 4 x + 4 touches the x-axis at one point.
Let's plot the function
.
The graph of this function is a parabola. It touches the x-axis (axis) at one point:
.
This point is the root of the original equation (2.1). Because this root is factored twice:
,
then such a root is usually called a multiple. That is, they believe that there are two equal roots:
.
Answer
;
.
Example 3
Find the roots of a quadratic equation:
(3.1)
.
Solution
Let's write the quadratic equation in general form:
(1)
.
Let's rewrite the original equation (3.1):
.
Comparing with (1), we find the values of the coefficients:
.
We find the discriminant:
.
The discriminant is negative, . Therefore there are no real roots.
You can find complex roots:
;
;
Let's plot the function
.
The graph of this function is a parabola. It does not intersect the x-axis (axis). Therefore there are no real roots.
Answer
There are no real roots. Complex roots:
;
;
.
Let's find the sum and product of the roots of the quadratic equation. Using formulas (59.8) for the roots of the above equation, we obtain
(the first equality is obvious, the second is obtained after a simple calculation, which the reader will carry out independently; it is convenient to use the formula for multiplying the sum of two numbers by their difference).
The following has been proven
Vieta's theorem. The sum of the roots of the above quadratic equation is equal to the second coefficient with the opposite sign, and their product is equal to the free term.
In the case of an unreduced quadratic equation, one should substitute the expressions of formula (60.1) into formulas (60.1) and take the form
Example 1. Compose a quadratic equation using its roots:
Solution, a) We find the equation has the form
Example 2. Find the sum of the squares of the roots of the equation without solving the equation itself.
Solution. The sum and product of the roots are known. Let us represent the sum of squared roots in the form
and we get
From Vieta's formulas it is easy to obtain the formula
expressing the rule for factoring a quadratic trinomial.
Indeed, let us write formulas (60.2) in the form
Now we have
which is what we needed to get.
The above derivation of Vieta's formulas is familiar to the reader from a high school algebra course. Another conclusion can be given using Bezout’s theorem and factorization of the polynomial (paragraphs 51, 52).
Let the roots of the equation then, according to the general rule (52.2), the trinomial on the left side of the equation is factorized:
Opening the parentheses on the right side of this identical equality, we obtain
and comparing the coefficients at the same powers will give us the Vieta formula (60.1).
The advantage of this derivation is that it can be applied to equations of higher degrees in order to obtain expressions for the coefficients of the equation in terms of its roots (without finding the roots themselves!). For example, if the roots of the given cubic equation
the essence is that according to equality (52.2) we find
(in our case, opening the brackets on the right side of the equality and collecting the coefficients at various degrees, we get
When solving arithmetic and algebraic problems, it is sometimes necessary to construct fraction V square. The easiest way to do this is when fraction decimal - a regular calculator is enough. However, if fraction ordinary or mixed, then when raising such a number to square Some difficulties may arise.
You will need
- calculator, computer, Excel application.
Instructions
To raise a decimal fraction V square, take an engineering one, type on it what is being built in square fraction and press the raise to the second power key. On most calculators this button is labeled "x²". On a standard Windows calculator, the function of raising to square looks like "x^2". For example, square the decimal fraction 3.14 will be equal to: 3.14² = 9.8596.
To build into square decimal fraction on a regular (accounting) calculator, multiply this number by itself. By the way, some models of calculators provide the ability to raise a number to square even in the absence of a special button. Therefore, first read the instructions for your specific calculator. Sometimes "tricky" exponentiations are given on the back cover or on the calculator. For example, on many calculators, to raise a number to square Just press the “x” and “=” buttons.
For construction in square common fraction (consisting of a numerator and a denominator), raise to square separately the numerator and denominator of this fraction. That is, use the following rule: (h / z)² = h² / z², where h is the numerator of the fraction, z is the denominator of the fraction. Example: (3/4)² = 3²/4² = 9/16.
If being built in square fraction– mixed (consists of an integer part and an ordinary fraction), then first reduce it to an ordinary form. That is, apply the following formula: (c h/z)² = ((c*z+ch) / z)² = (c*z+ch)² / z², where c is the integer part of the mixed fraction. Example: (3 2/5)² = ((3*5+2) / 5)² = (3*5+2)² / 5² = 17² / 5² = 289/25 = 11 14/25.
If in square(not ) fractions happen all the time, then use MS Excel. To do this, enter the following formula into one of the tables: = DEGREE (A2;2) where A2 is the address of the cell into which the raised value will be entered square fraction.To tell the program that the input number should be treated as fraction yu (i.e. do not convert it to decimal), type before fraction I have the number “0” and the sign “space”. That is, to enter, for example, the fraction 2/3, you need to enter: “0 2/3” (and press Enter). In this case, the decimal representation of the entered fraction will be displayed in the input line. The value and representation of the fraction itself will be saved in its original form. In addition, when using mathematical functions whose arguments are ordinary fractions, the result will also be presented as an ordinary fraction. Hence square the fraction 2/3 will be represented as 4/9.