The study of a polynomial of the second degree has received much attention in the course of algebra of the eighth grade. If this material is poorly understood by the schoolchild, then problems are unavoidable at the examinations of the exam and exam, both at the relevant level and at the base. Mandatory skills related to quadratic functions include plotting and analyzing graphs, solving equations.
Factoring a square trinomial is one of the standard school tasks. It is auxiliary in solving inequality by the interval method.
Finding the roots of an equation
The first thing necessary to factor the polynomial is to find its roots.
Roots are numbers that turn the sum of monomials in a polynomial to zero, which graphically looks like an intersection with a horizontal axis. They are determined using the discriminant or the Vieta theorem.
The discriminant of the trinomial ax 2 + bx + c is calculated by the formula: D = b 2t - 4ac.
In the case when the discriminant is not negative, the roots are expressed through it and the coefficients of the polynomial:
x 1 = 1/2 (-b + √D); x 2 = 1/2 (-b - √D)
If the discriminant is zero, x 1 and x 2 match.
To solve some trinomials, it is convenient to use the Vieta theorem:
x 1 + x 2 = -b: a; x 1 × x 2 = c: a
A certain mathematical intuition is required to apply the theorem. The bottom line is, knowing the sum and product of two unknowns, select these numbers. If they exist, then they are uniquely (up to a permutation).
One can verify the validity of the theorem by calculating the sum and product of roots in a general form. The formulas for x 1 and x 2 are also verified by direct substitution.
Factor Rule
The problem can be solved in real numbers if the polynomial has roots. The decomposition is determined by the formula:
ax 2 + bx + c = a (x - x 1 ) (x - x 2 )
Examples
Problem: Find the factorization of square trinomials.
a) x 2 - 6x + 5
Solution: we write out the coefficients of the trinomial:
a = 1; b = -6; c = 5.
We use the Vieta theorem:
x 1 + x 2 = 6;
x 1 × x 2 = 5.
It can be seen that x 1 = 1, x 2 = 5.
If it is not possible to quickly find the roots according to the equalities of the theorem, it is worth immediately proceeding to the calculation of the discriminant.
After the roots are found, you need to substitute them in the formula for decomposition:
x 2 - 6x + 5 = (x - 1) (x - 5)
The result recorded in this form can be considered final.
b) 2x 2 + x - 1
Decision:
a = 2, b = 1, c = -1.
If the leading coefficient is different from 1, the application of the Vieta theorem usually requires more time than the solution through the discriminant, so we proceed to its calculation.
D = 1 - 4 × 2 × (-1) = 9.
x 1 = 1/2; x 2 = -1.
By the formula it turns out:
2x 2 + x - 1 = 2 (x - 1/2) (x + 1).
c) x 2 - 8x + 16
Decision:
a = 1; b = -8; c = 16.
D = 0.
Since the discriminant is equal to zero, we have the case of coincidence of roots:
x 1 = x 2 = 4.
This situation, however, does not fundamentally differ from those considered previously.
x 2 - 8x + 16 = 1 (x - 4) (x - 4)
The result is often written in the form: (x - 4) 2 .
d) x 2 - 7x + 1
Decision:
a = 1; b = -7; c = 1.
D = 45.
The given example differs from the previous ones in that a rational root cannot be extracted from the discriminant. Hence, the roots of the polynomial are irrational.
x 1 = -1/2 (7 + √45); x 2 = -1/2 (7 - √45).
Or, the same thing
x 1 = -3.5 - 1 / 2√45; x 2 = -3.5 + 1 / 2√45.
The latter option is more convenient to use for recording decomposition. Omitting the leading coefficient equal to 1 here, we get:
x 2 - 7x + 1 = (x + 3.5 + 1 / 2√45) (x + 3.5 - 1 / 2√45)
For the case when the discriminant is negative, the following answer is enough within the framework of the school curriculum: the trinomial has no roots and, therefore, is not factorized. Such trinomials are also called irreducible. It is important to understand that we are only talking about the presence or absence of valid roots.
If the field of complex numbers is considered, factorization of the quadratic trinomial is possible for any discriminant.
Common mistakes
1) Many at the very beginning of the study of the polynomial incorrectly write out the coefficients, for example, pay attention to the order of monomials in the record.
So, the highest coefficient a in equation 101 - 79x + 38x 2 is 38, not 101, as you might think.
Another mistake related to the coefficients of the equation is the so-called "sign loss". In the same example, the coefficient b = -79, not 79.
2) Getting used to using the Vieta theorem for the case when a = 1, schoolchildren sometimes forget about its full formulation. In the polynomial from the first paragraph, it is incorrect to assume that the sum of the roots is 79, since the first coefficient is different from 1.
3) Computational errors - the most common problem of students. In many cases, substitution checking helps to avoid them.
Polynomials of the third degree and higher
Higher polynomials are rarely considered at school, since the task of finding the roots for polynomials of the third degree and higher is laborious. There are algorithms of high computational complexity for decomposing a polynomial of the third and fourth degree. For the fifth degree and above, a theorem on the unsolvability of an equation in radicals in a general form is proved.
Particular cases of these polynomials, which can be considered in high school, are characterized by the presence of rational easily selected roots. The number of the latter cannot exceed the degree of the polynomial. When working with the complex plane, their number exactly coincides with the highest degree.
Odd degree polynomials always have at least one real root. This is easy to show graphically - the continuous function defined by such a polynomial has both positive and negative values, which means it passes through 0.
All roots of two polynomials coincide if and only if their coefficients are proportional.
In general, the problem of finding roots and the problem of constructing a decomposition can be considered equivalent.