Introduction
Factorising is a cornerstone skill in algebra that turns a polynomial into a product of simpler expressions. When you see an expression like 2x² + 7x + 15, the goal is to rewrite it as a multiplication of two binomials (or a binomial and a constant) so that you can solve equations, simplify fractions, or analyse the behaviour of functions. In this article we will factorise 2x² + 7x + 15 step by step, explore the theory behind the method, look at real‑world examples, debunk common misconceptions, and answer the most frequently asked questions. By the end, you’ll have a clear, practical understanding of how to factorise this quadratic and similar expressions with confidence.
Detailed Explanation
A quadratic polynomial of the form ax² + bx + c can often be expressed as a product of two binomials:
[
(ax + m)(bx + n) = a x^2 + (an + bm) x + mn
]
where (m) and (n) are numbers that satisfy two conditions:
- (m \times n = c) (their product equals the constant term)
- (a \times n + b \times m = b) (their weighted sum equals the coefficient of (x))
When (a = 1), the process is straightforward: you simply find two numbers that multiply to (c) and add to (b). The standard approach is to multiply (a) and (c) first, then find two numbers that multiply to that product and add to (b). Even so, when (a) is not 1—as in 2x² + 7x + 15—you must account for the leading coefficient. Those two numbers become the “cross‑terms” in the factorisation.
It sounds simple, but the gap is usually here.
For 2x² + 7x + 15, we multiply (a) and (c):
(2 \times 15 = 30).
Those integers are 5 and 6. We now look for two integers whose product is 30 and whose sum is 7. In real terms, this observation guides the factorisation:
[
2x^2 + 7x + 15 = 2x^2 + 5x + 2x + 15
]
Next, we group the terms:
[
(2x^2 + 5x) + (2x + 15)
]
Factor out the greatest common factor (GCF) from each group:
[
x(2x + 5) + 1(2x + 5)
]
Since both groups contain the common binomial ((2x + 5)), we factor it out:
[
(2x + 5)(x + 3)
]
Thus, 2x² + 7x + 15 = (2x + 5)(x + 3). This factorisation is verified by expanding the product, which returns the original quadratic.
Step‑by‑Step Breakdown
- Identify the coefficients: (a = 2), (b = 7), (c = 15).
- Multiply (a) and (c): (2 \times 15 = 30).
- Find two numbers that multiply to 30 and add to 7: 5 and 6.
- Rewrite the middle term using these numbers:
[ 2x^2 + 5x + 2x + 15 ] - Group the terms:
[ (2x^2 + 5x) + (2x + 15) ] - Factor each group:
[ x(2x + 5) + 1(2x + 5) ] - Factor out the common binomial:
[ (2x + 5)(x + 3) ] - Check by expansion to ensure the product equals the original polynomial.
This systematic approach works for any quadratic where (a \neq 1) and the polynomial is factorable over the integers.
Real Examples
1. Solving a Quadratic Equation
Suppose you need to solve (2x^2 + 7x + 15 = 0).
After factorising, you get ((2x + 5)(x + 3) = 0).
Setting each factor to zero gives the solutions:
[
2x + 5 = 0 ;\Rightarrow; x = -\frac{5}{2} \
x + 3 = 0 ;\Rightarrow; x = -3
]
Thus, the equation has two real roots, (-\frac{5}{2}) and (-3) It's one of those things that adds up. Nothing fancy..
2. Simplifying a Rational Expression
Consider (\frac{2x^2 + 7x + 15}{x + 3}).
Factorising the numerator yields (\frac{(2x + 5)(x + 3)}{x + 3}).
The ((x + 3)) terms cancel (provided (x \neq -3)), leaving (2x + 5).
This simplification is useful in calculus when evaluating limits or in algebraic manipulation.
3. Graphing a Parabola
The quadratic (y = 2x^2 + 7x + 15) can be rewritten as (y = (2x + 5)(x + 3)).
The roots (-\frac{5}{2}) and (-3) are the x‑intercepts of the parabola.
Knowing the factorisation allows you to quickly sketch the graph and identify key points such as the vertex and axis of symmetry That's the whole idea..
Scientific or Theoretical Perspective
The factorisation of quadratics is rooted in the Fundamental Theorem of Algebra, which states that a polynomial of degree (n) has exactly (n) roots (counting multiplicities) in the complex number system. For a quadratic, there are always two roots, real or complex. Factorising over the integers is essentially finding the rational roots that satisfy the polynomial.
The AC method (multiply (a) and (c) and split the middle term) is a practical application of the distributive property and the concept of common factors. It leverages the fact that any integer factorisation of a polynomial must preserve both the product of the constant terms and the sum of the cross‑terms.
When the quadratic is not factorable over the integers, we turn to the Quadratic Formula:
