Problem 79 - Elementary Algebra

Question

Create a trinomial of the form \(a x_{2}+b x+c\) that does not factor and share it along with the reason why it does not factor.

Step-by-Step Solution

Answer

The trinomial \(5x^2 - 6x + 2\) does not factor because its discriminant \((-4)\) is negative.

Step 1: Choose Values for a, b, and c

Select integer values for \(a\), \(b\), and \(c\) such that the trinomial does not factor neatly over the integers. Let's choose \(a = 5\), \(b = -6\), and \(c = 2\). So, the trinomial is \(5x^2 - 6x + 2\).

Step 2: Confirm Non-factorability Using the Discriminant

For a trinomial \(ax^2 + bx + c\), check if it can be factored over the integers by using the discriminant: \(b^2 - 4ac\). Substitute \(a = 5\), \(b = -6\), and \(c = 2\): \((-6)^2 - 4 \times 5 \times 2 = 36 - 40 = -4\).

Step 3: Interpret the Discriminant Result

Since the discriminant \((-4)\) is negative, there are no real roots. Hence, the trinomial does not factor over the integers.

Key Concepts

Discriminant Method

When dealing with trinomials, one effective way to determine if a trinomial is factorable is by using the discriminant method. This involves the formula:

• For a quadratic equation of the form \( ax^2 + bx + c \), the discriminant \( \Delta \) is defined as \( b^2 - 4ac \).

The discriminant can tell us about the nature of the roots of the equation. Here's how you can interpret it:

• If \( \Delta > 0 \), the equation has two distinct real roots, indicating that it can be factored over the real numbers.
• If \( \Delta = 0 \), there is exactly one real root (a repeated root), and the trinomial can still be factored as a perfect square.
• If \( \Delta < 0 \), the equation has no real roots, which means it cannot be factored over the integers.

For example, in the trinomial \( 5x^2 - 6x + 2 \), we calculated the discriminant to be \(-4\). This negative discriminant confirms that no real roots exist, hence the trinomial cannot be neatly factored over the integers.

Integer Coefficients

In the realm of polynomial equations, the coefficients play a significant role. For a polynomial to be considered over the integers, all of its coefficients \( a \), \( b \), and \( c \) must be integers themselves. When performing operations like factoring, it's crucial to ensure these coefficients remain integers if we want the factorization to be valid over the set of integers.

• Integer coefficients simplify arithmetic operations like addition, subtraction, and multiplication, as these operations result in integer outputs.
• Within the context of factorization, a common goal is to express a polynomial as a product of other polynomials with integer coefficients.

By selecting integers like \( a = 5 \), \( b = -6 \), and \( c = 2 \) for the trinomial \( 5x^2 - 6x + 2 \), we ensure every step of checking factorability aligns with integer arithmetic, a fundamental aspect when dealing with such problems.

Non-factorable Polynomials

Non-factorable polynomials, sometimes known as irreducible polynomials, are those that cannot be expressed as a product of two or more non-trivial polynomials with integer coefficients. This property means that within the context of integer factorization, the polynomial is "prime."

• For a quadratic like \( ax^2 + bx + c \), if the discriminant \( b^2 - 4ac \) is less than zero, it suggests real solutions are nonexistent, making the trinomial non-factorable over the integers.
• Non-factorable polynomials pose challenges for solving equations as traditional factorization techniques do not apply. Instead, numeric or more advanced algebraic methods must be employed.

Take \( 5x^2 - 6x + 2 \) as a classic instance. Here, the discriminant is \(-4\), which leads to complex roots, confirming our trinomial remains non-factorable over the integers. Understanding this property helps significantly in algebraic problem-solving, as it delineates the limits of simpler solution methods.