Online calculator.
Isolating the square of a binomial and factoring a square trinomial.
This math program distinguishes the square binomial from the square trinomial, i.e. does a transformation like: \(ax^2+bx+c \rightarrow a(x+p)^2+q \) and factorizes quadratic trinomial : \(ax^2+bx+c \rightarrow a(x+n)(x+m) \)
Those. the problems boil down to finding the numbers \(p, q\) and \(n, m\)
The program not only gives the answer to the problem, but also displays the solution process.
This program may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.
If you are not familiar with the rules for entering a quadratic trinomial, we recommend that you familiarize yourself with them.
Rules for entering a quadratic polynomial
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.
Numbers can be entered as whole or fractional numbers.
Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
Rules for entering decimal fractions.
In decimal fractions, the fractional part can be separated from the whole part by either a period or a comma.
For example, you can enter decimals like this: 2.5x - 3.5x^2
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &
Input: 3&1/3 - 5&6/5x +1/7x^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) x + \frac(1)(7)x^2\)
When entering an expression you can use parentheses. In this case, when solving, the introduced expression is first simplified.
For example: 1/2(x-1)(x+1)-(5x-10&1/2)
Example detailed solution
Isolating the square of a binomial.$$ ax^2+bx+c \rightarrow a(x+p)^2+q $$ $$2x^2+2x-4 = $$ $$2x^2 +2 \cdot 2 \cdot\left( \frac(1)(2) \right)\cdot x+2 \cdot \left(\frac(1)(2) \right)^2-\frac(9)(2) = $$ $$2\left (x^2 + 2 \cdot\left(\frac(1)(2) \right)\cdot x + \left(\frac(1)(2) \right)^2 \right)-\frac(9 )(2) = $$ $$2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Answer:$$2x^2+2x-4 = 2\left(x+\frac(1)(2) \right)^2-\frac(9)(2) $$ Factorization.$$ ax^2+bx+c \rightarrow a(x+n)(x+m) $$ $$2x^2+2x-4 = $$
$$ 2\left(x^2+x-2 \right) = $$
$$ 2 \left(x^2+2x-1x-1 \cdot 2 \right) = $$ $$ 2 \left(x \left(x +2 \right) -1 \left(x +2 \right ) \right) = $$ $$ 2 \left(x -1 \right) \left(x +2 \right) $$ Answer:$$2x^2+2x-4 = 2 \left(x -1 \right) \left(x +2 \right) $$
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A little theory.
Isolating the square of a binomial from a square trinomial
If the square trinomial ax 2 +bx+c is represented as a(x+p) 2 +q, where p and q are real numbers, then we say that from square trinomial, the square of the binomial is highlighted.
From the trinomial 2x 2 +12x+14 we extract the square of the binomial.
\(2x^2+12x+14 = 2(x^2+6x+7) \)
To do this, imagine 6x as a product of 2*3*x, and then add and subtract 3 2. We get:
$$ 2(x^2+2 \cdot 3 \cdot x + 3^2-3^2+7) = 2((x+3)^2-3^2+7) = $$ $$ = 2 ((x+3)^2-2) = 2(x+3)^2-4 $$
That. We extract the square binomial from the square trinomial, and showed that:
$$ 2x^2+12x+14 = 2(x+3)^2-4 $$
Factoring a quadratic trinomial
If the square trinomial ax 2 +bx+c is represented in the form a(x+n)(x+m), where n and m are real numbers, then the operation is said to have been performed factorization of a quadratic trinomial.
Let us show with an example how this transformation is done.
Let's factor the quadratic trinomial 2x 2 +4x-6.
Let us take the coefficient a out of brackets, i.e. 2:
\(2x^2+4x-6 = 2(x^2+2x-3) \)
Let's transform the expression in brackets.
To do this, imagine 2x as the difference 3x-1x, and -3 as -1*3. We get:
$$ = 2(x^2+3 \cdot x -1 \cdot x -1 \cdot 3) = 2(x(x+3)-1 \cdot (x+3)) = $$
$$ = 2(x-1)(x+3) $$
That. We factored the quadratic trinomial, and showed that:
$$ 2x^2+4x-6 = 2(x-1)(x+3) $$
Note that factoring a quadratic trinomial is possible only if the quadratic equation corresponding to this trinomial has roots.
Those. in our case, it is possible to factor the trinomial 2x 2 +4x-6 if the quadratic equation 2x 2 +4x-6 =0 has roots. In the process of factorization, we established that the equation 2x 2 + 4x-6 = 0 has two roots 1 and -3, because with these values, the equation 2(x-1)(x+3)=0 turns into a true equality.
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 very to a large extent– 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.
Formulas for different meanings discriminants differ.
If D is positive:
If D is zero:
Online calculators
On the Internet there is online calculator. 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 invite you to view simple examples, 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 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. Fits last option. 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.
Factoring quadratic trinomials is one of the school assignments that everyone faces sooner or later. How to do it? What is the formula for factoring a quadratic trinomial? Let's figure it out step by step with the help of examples.
General formula
The factorization of square trinomials is carried out by solving quadratic equation. This is a simple problem that can be solved by several methods - by finding the discriminant, using Vieta's theorem, there is also a graphical solution. The first two methods are studied in high school.
The general formula looks like this:lx 2 +kx+n=l(x-x 1)(x-x 2) (1)
Algorithm for completing the task
In order to factor quadratic trinomials, you need to know Vita's theorem, have a solution program at hand, be able to find a solution graphically, or look for roots of a second-degree equation using the discriminant formula. If a quadratic trinomial is given and it needs to be factorized, the algorithm is as follows:
1) Equate the original expression to zero to obtain an equation.
2) Give similar terms (if necessary).
3) Find the roots using any known method. The graphical method is best used if it is known in advance that the roots are integers and small numbers. It must be remembered that the number of roots is equal to the maximum degree of the equation, that is, the quadratic equation has two roots.
4) Substitute the value X into expression (1).
5) Write down the factorization of quadratic trinomials.
Examples
Practice allows you to finally understand how this task is performed. The following examples illustrate the factorization of a quadratic trinomial:
it is necessary to expand the expression:
Let's resort to our algorithm:
1) x 2 -17x+32=0
2) similar terms are reduced
3) using Vieta’s formula, it is difficult to find roots for this example, so it is better to use the expression for the discriminant:
D=289-128=161=(12.69) 2
4) Let’s substitute the roots we found into the basic formula for decomposition:
(x-2.155) * (x-14.845)
5) Then the answer will be like this:
x 2 -17x+32=(x-2.155)(x-14.845)
Let's check whether the solutions found by the discriminant correspond to the Vieta formulas:
14,845 . 2,155=32
For these roots, Vieta’s theorem is applied, they were found correctly, which means the factorization we obtained is also correct.
Similarly, we expand 12x 2 + 7x-6.
x 1 =-7+(337) 1/2
x 2 =-7-(337)1/2
In the previous case, the solutions were non-integer, but real numbers, which are easy to find if you have a calculator in front of you. Now let's look at more complex example, in which the roots will be complex: factor x 2 + 4x + 9. Using Vieta's formula, the roots cannot be found, and the discriminant is negative. The roots will be on the complex plane.
D=-20
Based on this, we obtain the roots that interest us -4+2i*5 1/2 and -4-2i * 5 1/2 since (-20) 1/2 = 2i*5 1/2.
We obtain the desired decomposition by substituting the roots into the general formula.
Another example: you need to factor the expression 23x 2 -14x+7.
We have the equation 23x 2 -14x+7 =0
D=-448
This means the roots are 14+21.166i and 14-21.166i. The answer will be:
23x 2 -14x+7 =23(x- 14-21,166i )*(X- 14+21,166i ).
Let us give an example that can be solved without the help of a discriminant.
Let's say we need to expand the quadratic equation x 2 -32x+255. Obviously, it can also be solved using a discriminant, but in this case it is faster to find the roots.
x 1 =15
x 2 =17
Means x 2 -32x+255 =(x-15)(x-17).
Square trinomial is called a trinomial of the form a*x 2 +b*x+c, where a,b,c are some arbitrary real numbers, and x is a variable. Moreover, the number a should not be equal to zero.
The numbers a,b,c are called coefficients. The number a is called the leading coefficient, the number b is the coefficient of x, and the number c is called the free term.
Root of a square trinomial a*x 2 +b*x+c is any value of the variable x such that the square trinomial a*x 2 +b*x+c vanishes.
In order to find the roots of a quadratic trinomial, it is necessary to solve a quadratic equation of the form a*x 2 +b*x+c=0.
How to find the roots of a quadratic trinomial
To solve this, you can use one of the known methods.
- 1 way.
Finding the roots of a square trinomial using the formula.
1. Find the value of the discriminant using the formula D =b 2 -4*a*c.
2. Depending on the value of the discriminant, calculate the roots using the formulas:
If D > 0, then the square trinomial has two roots.
x = -b±√D / 2*a
If D< 0, then the square trinomial has one root.
If the discriminant is negative, then the quadratic trinomial has no roots.
- Method 2.
Finding the roots of a quadratic trinomial by isolating full square. Let's look at the example of the given quadratic trinomial. A reduced quadratic equation whose leading coefficient is equal to one.
Let's find the roots of the quadratic trinomial x 2 +2*x-3. To do this, we solve the following quadratic equation: x 2 +2*x-3=0;
Let's transform this equation:
On the left side of the equation there is a polynomial x 2 +2*x, in order to represent it as a square of the sum we need there to be another coefficient equal to 1. Add and subtract 1 from this expression, we get:
(x 2 +2*x+1) -1=3
What can be represented in parentheses as the square of a binomial
This equation breaks down into two cases: either x+1=2 or x+1=-2.
In the first case, we get the answer x=1, and in the second, x=-3.
Answer: x=1, x=-3.
As a result of the transformations, we need to get the square of the binomial on the left side, and a certain number on the right side. The right side should not contain a variable.
You can find the root of a square trinomial using the discriminant. In addition, for the reduced polynomial of the second degree, Vieta’s theorem, based on the ratio of the coefficients, applies.
Instructions
- Quadratic equations are a fairly extensive topic in school algebra. The left side of such an equation is a polynomial of the second degree of the form A x² + B x + C, i.e. an expression of three monomials of varying degrees of unknown x. To find the root of a square trinomial, you need to calculate the value of x at which this expression is equal to zero.
- To solve a quadratic equation, you need to find the discriminant. Its formula is a consequence of isolating the complete square of the polynomial and represents a certain ratio of its coefficients: D = B² – 4 A C.
- The discriminant can take different meanings, including being negative. And if younger schoolchildren can say with relief that such an equation has no roots, then high school students are already able to determine them based on the theory of complex numbers. So, there can be three options: Discriminant – a positive number. Then the roots of the equation are equal: x1 = (-B + √D)/2 A; x2 = (-B - √D)/2 A;
The discriminant went to zero. Theoretically, in this case the equation also has two roots, but practically they are the same: x1 = x2 = -B/2 A;
The discriminant is less than zero. A certain value i² = -1 is introduced into the calculation, which allows us to write comprehensive solution: x1 = (-B + i √|D|)/2 A; x2 = (-B - i √|D|)/2 A. - The discriminant method is valid for any quadratic equation, but there are situations when it is advisable to use more quick way, especially for small integer coefficients. This method is called Vieta’s theorem and consists of a pair of relationships between the coefficients in the reduced trinomial: x² + P x + Q
x1 + x2 = -P;
x1 x2 = Q. All that remains is to find the roots. - It should be noted that the equation can be reduced to a similar form. To do this, you need to divide all the terms of the trinomial by the coefficient of the highest power A: A x² + B x + C |A
x² + B/A x + C/A
x1 + x2 = -B/A;
x1 x2 = C/A.