# Trinomial Forms x ^ 2 + bx + c (with Examples)

Before learning to solve trinomials from the form x ^ 2 + bx + c , and before even knowing the trinomial concept, it is important to know two important concepts; that is, the concepts of monomial and polynomial. Monomial is an expression of type a * x n , where a is a rational number, n is a natural number and x is a variable.

A polynomial is a linear combination of the monomial forms a n * x n + a n-1 * x n-1 + … + a 2 * x 2 + a 1 * x + a 0 , where each a i , with i = 0, …, n, are rational numbers, n is a natural number and a_n is zero. In this case it is said that the degree of polynomial is n.

The polynomial formed by the sum of only two terms (two monomials) with different degrees, is known as a binomial.

Index

• 1 Trinomial
• 1.1 Trinomial square is perfect
• 2 Characteristics of class 2 trinomials
• 2.1 Perfect box
• 2.2 Solvent formulas
• 2.3 Geometric interpretation
• 2.4 Factoring from trinomials
• 3 Example
• 3.1 Example 1
• 3.2 Example 2
• 4 References

## Trinomi

A polynomial formed by the sum of only three terms (three monomials) with different degrees is known as a trinomial. Following is an example of trinomial:

• 3 + x 2 + 5x
• 2x 4 -x 3 + 5
• 2 + 6x + 3

There are several types of trinomials. All of this highlights the perfect square trinomial.

### Trinomial square is perfect

Perfect Trinomial squared is the result of the appointment of binomial squared. As an example:

• (3x-2) 2 = 9x 2 -12x + 4
• (2x 3 + y) 2 = 4x 6 + 4x 3 y + y 2
• (4x 2 -2 years 4 ) 2 = 16x 4 -16x 2 and 4 + 4thn 8
• 1 / 16x 2 and 8 -1 / 2xy 4 z + z 2 = (1 / 4xy 4 ) 2 -2 (1 / 4xy 4 ) z + z 2 = (1 / 4xy 4 -z) 2

## Class 2 trinomial characteristics

### Perfect box

In general, the trinomial of the ax 2 + bx + c is a perfect square if the discriminant is zero; that is, if b 2 -4ac = 0, because in this case it will only have one root and can be expressed in the form a (xd) 2 = (√a (xd)) 2 , where d is the root already mentioned.

The polynomial root is a number where the polynomial becomes zero; in other words, a number that, by replacing it with x in the polynomial expression, produces zero.

### Solvent Formula

The general formula for calculating the second degree polynomial root of the 2 + bx + c form ax is the resolver formula, which states that this root is given by (-b ± √ (b 2 -4ac)) / 2a, where b 2 -4ac is known as discriminant and usually denoted by Δ. From this formula following the ax 2 + bx + c has:

– Two different real roots if Δ> 0.

– Original single root if Δ = 0.

– Has no real root if Δ <0.

In the following we will consider only trinomials of the form x 2 + bx + c, where obviously c must be a nonzero number (otherwise it will be binomial). This type of Trinomial has certain advantages when factoring and operating it.

### Geometric Interpretation

Geometrically, trinomial x 2 + bx + c is a parabola that opens upward and has a point at the point (-b / 2, -b 2 /4 + c) of the Cartesian plane as x 2 + bx + c = (x + b / 2) 2 -b 2 /4 + c.

This parabola crosses the Y axis at points (0, c) and the X axis at points (d 1 , 0) and (d) 2 , 0); then, d 1 and d 2 they are the roots of the trinomial. It can happen that the trinomial has one d root, in this case the only chunk with the X axis is (d, 0).

It could also happen that the trinomial has no real roots, in this case it will not cut the X axis at any point.

For example, x 2 + 6x + 9 = (x + 3) 2 -9 + 9 = (x + 3) 2 is a parabolic node at (-3.0), which intersects the Y axis at (0.9) and the axis X at (-3.0). ### Trinomial Factorization

A very useful tool when working with polynomials is factoring, which is to express polynomials as a product of factors. In general, given trinomial in the form of x 2 + bx + c, does this have two different roots d 1 and d 2 , can be factored as (xd) 1 ) (xd) 2 ).

If you only have one root d, you can factor it as (xd) (xd) = (xd) 2 , and if it doesn’t have a real root, it remains the same; in this case it does not support factorization as a product of factors other than itself.

This means that, knowing the roots of a trinomial from an established form, factorization can be easily expressed, and as already stated, these roots can always be determined using solutions.

However, there are a large number of types of trinomy that can be factored without having to know the roots beforehand, which simplifies the work.

The root can be determined directly from factorization without the need to use the resolver formula; this is the polynomial of the form x 2 + (a + b) x + ab. In this case you have:

2 + (a + b) x + ab = x 2 + ax + bx + ab = x (x + a) + b (x + a) = (x + b) (x + a).

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

In other words, given trinomial x 2 + bx + c, if there are two numbers u and v such that c = uv and b = u + v, then x 2 + bx + c = (x + u) (x + v).

That is, given a trinomial x 2 + bx + c, first verify whether there are two numbers multiplied by the independent term (c) and added (or subtracted, depending on the case), give the term that accompanies x (b).

Not with all trinomials in this way this method can be applied; where you can’t, you go to resolvent and apply the above mentioned.

## Example

### Example 1

For the following trinomial factors x 2 + 3x + 2 we proceed as follows:

You have to find two numbers so that when you add it, the result is 3, and when you multiply it, the result is 2.

After conducting an inspection it can be concluded that the numbers sought are: 2 and 1. Therefore, x 2 + 3x + 2 = (x + 2) (x + 1).

### Example 2

For the trinomial factor x 2 -5x + 6 we look for two numbers whose sum is -5 and the product is 6. The numbers that meet these two conditions are -3 and -2. Therefore, the factorization of the given trinomial is x 2 -5x + 6 = (x-3) (x-2 ##### byAbdullah Sam
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