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 factor the ill-fated trinomial is the theorem. She looks in the following way: “If x1 and x2 are roots quadratic trinomial ax^2+bx+c, then ax^2+bx+c=a(x-x1)(x-x2).” Of course, there is a proof of this theorem, but it requires some theoretical knowledge(when 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, therefore b/a=-(x1+x2), c/a=x1*x2. So, 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). , ax^2+bx+c=a(x-x1)(x-x2) Sometimes teachers force you to learn the proof, but if it is not required, I advise you to simply memorize the final formula.
Step 2
Let's take the trinomial 3x^2-24x+21 as an example. 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.
Step 3
Let's 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 instructions with 2 ways to solve them using the same equation as an example. The resulting roots are x1=7, x2=1.
Step 4
Now that we have the roots of the trinomial, 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 a term 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 resulting factors ((x-7), (3x-3) are polynomials of the first degree. That’s all the expansion =) If you doubt the answer received, you can always check it by multiplying the brackets.
Step 5
Checking 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 decision is correct! I hope my instructions will help someone =) Good luck with your studies!
- In our case, in the equation D > 0 and we got 2 roots. If there was a D<0, то уравнение, как и многочлен, соответственно, корней бы не имело.
- If a square trinomial has no roots, then it cannot be factorized, which are polynomials of the first degree.
Factoring a quadratic trinomial may be useful when solving inequalities from problem C3 or problem with parameter C5. Also, many B13 word problems will be solved much faster if you know Vieta’s theorem.
This theorem, of course, can be considered from the perspective of the 8th grade, in which it is taught for the first time. But our task is to prepare well for the Unified State Exam and learn to solve exam tasks as efficiently as possible. Therefore, this lesson considers an approach slightly different from the school one.
Formula for the roots of the equation using Vieta’s theorem Many people know (or at least have seen):
$$x_1+x_2 = -\frac(b)(a), \quad x_1 x_2 = \frac(c)(a),$$
where `a, b` and `c` are the coefficients of the quadratic trinomial `ax^2+bx+c`.
To learn how to easily use the theorem, let's understand where it comes from (this will actually make it easier to remember).
Let us have the equation `ax^2+ bx+ c = 0`. For further convenience, divide it by `a` and get `x^2+\frac(b)(a) x + \frac(c)(a) = 0`. Such an equation is called a reduced quadratic equation.
Important lesson idea: any quadratic polynomial that has roots can be expanded into parentheses. Let's assume that ours can be represented as `x^2+\frac(b)(a) x + \frac(c)(a) = (x + k)(x+l)`, where `k` and ` l` - some constants.
Let's see how the brackets open:
$$(x + k)(x+l) = x^2 + kx+ lx+kl = x^2 +(k+l)x+kl.$$
Thus, `k+l = \frac(b)(a), kl = \frac(c)(a)`.
This is slightly different from the classical interpretation Vieta's theorem- in it we look for the roots of the equation. I propose to look for terms for bracket decomposition- this way you don’t need to remember about the minus from the formula (meaning `x_1+x_2 = -\frac(b)(a)`). It is enough to select two such numbers, the sum of which is equal to the average coefficient, and the product is equal to the free term.
If we need a solution to the equation, then it is obvious: the roots `x=-k` or `x=-l` (since in these cases one of the brackets will be zero, which means the entire expression will be zero).
I'll show you the algorithm as an example: How to expand a quadratic polynomial into brackets.
Example one. Algorithm for factoring a quadratic trinomial
The path we have is a quadrant trinomial `x^2+5x+4`.
It is reduced (the coefficient of `x^2` is equal to one). He has roots. (To be sure, you can estimate the discriminant and make sure that it is greater than zero.)
Further steps (you need to learn them by completing all training tasks):
- Complete the following entry: $$x^2+5x+4=(x \ldots)(x \ldots).$$ Instead of dots, leave free space, we will add suitable numbers and signs there.
- Consider all possible options for decomposing the number `4` into the product of two numbers. We get pairs of “candidates” for the roots of the equation: `2, 2` and `1, 4`.
- Figure out which pair you can get the average coefficient from. Obviously it's `1, 4`.
- Write $$x^2+5x+4=(x \quad 4)(x \quad 1)$$.
- The next step is to place signs in front of the inserted numbers.
How to understand and forever remember what signs should appear before the numbers in brackets? Try opening them (brackets). The coefficient before `x` to the first power will be `(± 4 ± 1)` (we don’t know the signs yet - we need to choose), and it should be equal to `5`. Obviously, there will be two pluses $$x^2+5x+4=(x + 4)(x + 1)$$.
Perform this operation several times (hello, training tasks!) and you will never have any more problems with this.
If you need to solve the equation `x^2+5x+4`, then now solving it will not be difficult. Its roots are `-4, -1`.
Example two. Factorization of a quadratic trinomial with coefficients of different signs
Let us need to solve the equation `x^2-x-2=0`. Offhand, the discriminant is positive.
We follow the algorithm.
- $$x^2-x-2=(x \ldots) (x \ldots).$$
- There is only one factorization of two into integer factors: `2 · 1`.
- We skip the point - there is nothing to choose from.
- $$x^2-x-2=(x \quad 2) (x \quad 1).$$
- The product of our numbers is negative (`-2` is the free term), which means that one of them will be negative and the other will be positive.
Since their sum is equal to `-1` (the coefficient of `x`), then `2` will be negative (the intuitive explanation is that two is the larger of the two numbers, it will “pull” more strongly in the negative direction). We get $$x^2-x-2=(x - 2) (x + 1).$$
Third example. Factoring a quadratic trinomial
The equation is `x^2+5x -84 = 0`.
- $$x+ 5x-84=(x \ldots) (x \ldots).$$
- Decomposition of 84 into integer factors: `4 21, 6 14, 12 7, 2 42`.
- Since we need the difference (or sum) of the numbers to be 5, the pair `7, 12` is suitable.
- $$x+ 5x-84=(x\quad 12) (x\quad 7).$$
- $$x+ 5x-84=(x + 12) (x - 7).$$
Hope, expansion of this quadratic trinomial into brackets It's clear.
If you need a solution to an equation, here it is: `12, -7`.
Training tasks
I bring to your attention a few examples that are easy to are solved using Vieta's theorem.(Examples taken from the magazine "Mathematics", 2002.)
- `x^2+x-2=0`
- `x^2-x-2=0`
- `x^2+x-6=0`
- `x^2-x-6=0`
- `x^2+x-12=0`
- `x^2-x-12=0`
- `x^2+x-20=0`
- `x^2-x-20=0`
- `x^2+x-42=0`
- `x^2-x-42=0`
- `x^2+x-56=0`
- `x^2-x-56=0`
- `x^2+x-72=0`
- `x^2-x-72=0`
- `x^2+x-110=0`
- `x^2-x-110=0`
- `x^2+x-420=0`
- `x^2-x-420=0`
A couple of years after the article was written, a collection of 150 tasks for expanding a quadratic polynomial using Vieta’s theorem appeared.
Like and ask questions in the comments!
Square trinomial is called a polynomial of the form ax 2 +bx +c, Where x– variable, a,b,c– some numbers, and a ≠ 0.
Coefficient A called senior coefficient, c – free member square trinomial.
Examples of quadratic trinomials:
2 x 2 + 5x+4(Here a = 2, b = 5, c = 4)
x 2 – 7x + 5(Here a = 1, b = -7, c = 5)
9x 2 + 9x – 9(Here a = 9, b = 9, c = -9)
Coefficient b or coefficient c or both coefficients can be equal to zero at the same time. For example:
5 x 2 + 3x(Herea = 5,b = 3,c = 0, so there is no value for c in the equation).
6x 2 – 8 (Herea = 6, b = 0, c = -8)
2x2(Herea = 2, b = 0, c = 0)
The value of the variable at which the polynomial vanishes is called root of the polynomial.
To find the roots of a quadratic trinomialax 2 +
bx +
c, we need to equate it to zero -
that is, solve the quadratic equationax 2 +
bx +
c = 0 (see section "Quadratic equation").
Factoring a quadratic trinomial
Example:
Let's factorize the trinomial 2 x 2 + 7x – 4.
We see: coefficient A = 2.
Now let's find the roots of the trinomial. To do this, we equate it to zero and solve the equation
2x 2 + 7x – 4 = 0.
How to solve such an equation - see in the section “Formulas of the roots of a quadratic equation. Discriminant." Here we will immediately state the result of the calculations. Our trinomial has two roots:
x 1 = 1/2, x 2 = –4.
Let’s substitute the values of the roots into our formula, taking the value of the coefficient out of brackets A, and we get:
2x 2 + 7x – 4 = 2(x – 1/2) (x + 4).
The result obtained can be written differently by multiplying the coefficient 2 by the binomial x – 1/2:
2x 2 + 7x – 4 = (2x – 1) (x + 4).
The problem is solved: the trinomial is factorized.
Such an expansion can be obtained for any quadratic trinomial that has roots.
ATTENTION!
If the discriminant of a quadratic trinomial is zero, then this trinomial has one root, but when decomposing the trinomial, this root is taken as the value of two roots - that is, as the same value x 1 andx 2 .
For example, a trinomial has one root equal to 3. Then x 1 = 3, x 2 = 3.