# What are perfect square trinomials

## Trinomial of the form x ^ 2 + bx + c (with examples)

Before you learn to solve that Trinomial of the form x ^ 2 + bx + cand before knowing the concept of the trinomial, it is important to know two essential concepts; namely the concepts of monomial and polynomial. A monomial is an expression of the type a * xn, where a is a rational number, n is a natural number, and x is a variable.

A polynomial is a linear combination of monomials of the form an* xn+ an-1* xn-1+ ... + a2* x2+ a1* x + a0where everyoneI, with i = 0, ..., n, is a rational number, n is a natural number and a_n is not equal to zero. In this case, the degree of the polynomial is said to be n. A polynomial that is formed from the sum of just two terms (two monomials) of different degrees is known as a binomial.

index

• 1 trinome
• 1.1 Perfect square trinomial
• 2 characteristics of grade 2 trinomials
• 2.1 Perfect square
• 2.2 Solvent Formula
• 2.3 Geometric interpretation
• 2.4 Factorization of trinomials
• 3 examples
• 3.1 Example 1
• 3.2 Example 2
• 4 references

### Trinome

A polynomial that is formed from the sum of only three terms (three monomials) of different degrees is called a trinomial. The following are examples of trinomials:

• x3+ x2+ 5x
• 2x4-x3+5
• x2+ 6x + 3

There are different types of trinomials. Of these highlights, the perfect square trinomial.

### Perfect square trinomial

A perfect square trinomial is the result of squaring a binomial. For example:

• (3x-2)2= 9x2-12x + 4
• (2x3+ y)2= 4x6+ 4x3y + y2
• (4x2-2y4)2= 16x4-16x2and4+ 4Y8
• 1 / 16x2and8-1 / 2xy4z + z2= (1 / 4xy4)2-2 (1 / 4xy4) z + z2= (1 / 4xy4-z)2

### Perfect square

Usually a trinome of the form ax2+ bx + c is a perfect square if its discriminant is zero; that is, if b2-4ac = 0, since in this case it has only one root and can be expressed in the form a (x-d)2= (√a (x-d))2, where d is the root mentioned earlier.

A root of a polynomial is a number in which the polynomial becomes zero; in other words, a number which, by replacing it in x in the expression of the polynomial, results in zero.

### Solvent formula

A general formula for computing the roots of a second degree polynomial of the form ax2+ bx + c is the resolver's formula that says these roots are given by (-b ± √ (b2-4ac)) / 2a, where b2-4ac is known as a discriminant and is commonly referred to as Δ. From this formula it follows that ax2+ bx + c has:

- Two different real roots if Δ> 0.

- A single real root if Δ = 0.

- It has no real root if Δ <0.

In the following only the trinomials of the form x are considered2+ bx + c, where clearly c must be a number different from zero (otherwise it would be a binomial). These type of trinomials have certain advantages when factored and operated with them.

### Geometric interpretation

Geometrically, the trinomial is x2+ bx + c is a parabola that opens upwards and has the vertex at the point (-b / 2, -b2/ 4 + c) the Cartesian plane because x2+ bx + c = (x + b / 2)2-b2/ 4 + c.

This parabola intersects the Y axis at point (0, c) and the X axis at points (i.e.1, 0) and (d)2, 0); then d1 and d2 They are the roots of the trinomal. It can happen that the trinomial has a single root d, in which case the only intersection with the X axis would be (d, 0).

It can also happen that the trinome has no real roots, in which case it would not intersect the X-axis at any point.

Example: x2+ 6x + 9 = (x + 3)2-9 + 9 = (x + 3)2 is the parabola with the vertex at (-3.0), which intersects the Y-axis in (0.9) and the X-axis in (-3.0).

### Trinomial factorization

A very useful tool when working with polynomials is factoring, which is designed to express a polynomial as the product of factors. Usually given a trinomial of the form x2+ bx + c, if this has two different roots d1 and d2, can be factored as (x-d)1) (x-d)2).

If you only have one root d, you can write it as (x-d) (x-d) = (x-d)2and if it has no real roots it stays the same; In this case, it does not support factoring as the product of factors other than itself.

This means that knowing the roots of a trinomus of the already established form, its factorization can easily be expressed, and as mentioned earlier, these roots can always be determined using the resolver.

However, there is a significant amount of this type of trinomies that can be factored without first knowing their roots, which makes the job easier.

The roots can be determined directly from the factorization without having to use the resolver's formula; These are the polynomials of the form x2 + (a + b) x + from. In this case we have:

x2+ (a + b) x + ab = x2+ ax + bx + ab = x (x + a) + b (x + a) = (x + b) (x + a).

From here it is easily observed that the roots are -a and -b.

In other words, given a trinomial x2+ bx + c, if there are two numbers u and v such that c = uv and b = u + v, then x2+ bx + c = (x + u) (x + v).

That is, given a trinomial x2+ bx + c, first verify that there are two numbers multiplied by the independent expression (c) and added (or subtracted, depending on the case), give the expression that accompanies the x (b).

This method cannot be used with all trinomials in this way; where you can't, go to the solution and apply the above.

### example 1

To factor the following trinomial x2+ 3x + 2 we do the following:

You need to find two numbers so when you add them the result is 3 and when you multiply them the result is 2.

After inspection it can be concluded that the numbers we are looking for are: 2 and 1. Hence x2+ 3x + 2 = (x + 2) (x + 1).

### Example 2

To factorize the trinomial x2-5x + 6 searches for two numbers, the sum of which is -5 and the product of which is 6. The numbers that meet these two conditions are -3 and -2. Hence the factorization of the given trinomial is x2-5x + 6 = (x-3) (x-2).

### credentials

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