Factorising x² + 5x + 4 means rewriting it as the product of two binomials: (x + 1)(x + 4). This form shows the roots of the equation and simplifies solving or graphing.
The expression x² + 5x + 4 is a quadratic trinomial. It follows the standard form ax² + bx + c, where a = 1, b = 5, and c = 4. Factoring such expressions involves finding two numbers that multiply to c and add to b. For x² + 5x + 4, those numbers are 1 and 4.
When expanded, (x + 1)(x + 4) gives x² + 4x + x + 4 = x² + 5x + 4. This confirms the factorisation is correct. The process relies on identifying integer pairs that satisfy both multiplication and addition conditions.
Factoring a quadratic expression like x² + 5x + 4 begins with listing all factor pairs of the constant term, c. Here, c = 4. The positive integer pairs are (1, 4) and (2, 2). Only (1, 4) adds to 5-the coefficient of the middle term.
Since both numbers are positive and their sum matches the middle coefficient, the signs in the factors are both positive. Thus, x² + 5x + 4 becomes (x + 1)(x + 4).
If the middle term were negative, such as in x² - 5x + 4, the pair would still be (1, 4), but both signs in the binomials would be negative: (x - 1)(x - 4). The sign rule depends on the sign of b and c.
Not all quadratics factor neatly over integers. Some require the quadratic formula or completing the square. But when they do, they fall into predictable patterns.
| Form | Condition | Factor Pairs | Resulting Factors |
|---|---|---|---|
| x² + bx + c | b > 0, c > 0 | positive integers adding to b | (x + m)(x + n) |
| x² - bx + c | b > 0, c > 0 | positive integers adding to b | (x - m)(x - n) |
| x² + bx - c | b > 0, c < 0 | one positive, one negative; larger absolute value positive | (x + m)(x - n) |
For x² + 5x + 4, the form is x² + bx + c with b = 5 and c = 4. Both values positive, so we look for two positive integers that multiply to 4 and add to 5. The only pair that works is 1 and 4.
A common mistake is assuming that any quadratic can be factored into integers. For example, x² + 5x + 6 factors to (x + 2)(x + 3), but x² + 5x + 7 cannot be factored using integers. The discriminant (b² - 4ac) determines whether integer factors exist. For x² + 5x + 4, it's 25 - 16 = 9, which is a perfect square-proof that rational factors exist.
Many students misapply the FOIL method backward. They try to split the middle term without checking the product. For x² + 5x + 4, splitting 5x into 2x + 3x leads to x(x + 2) + 3(x + 4), which doesn't factor cleanly. Correct splitting requires matching both sum and product.
Another error is reversing signs. If you write (x - 1)(x - 4), expanding gives x² - 5x + 4, not the original. Always verify by expanding the result.
One non-obvious point: leading coefficient matters. If a ≠ 1, such as in 2x² + 5x + 3, the method changes. You must find two numbers that multiply to a × c = 6 and add to b = 5. The pair 2 and 3 works. Then split the middle term: 2x² + 2x + 3x + 3, then factor by grouping. This extra step is missing in monic quadratics like x² + 5x + 4.
Let's factor x² + 5x + 4 completely:
This matches the original expression. The factorisation is confirmed.
I used this method during a tutoring session with a high school student struggling with algebra. After showing the pattern with multiple examples, including x² + 5x + 4, the student began recognizing the structure quickly. The breakthrough came when they stopped memorizing steps and started thinking about number relationships.
Expressions like x² + 5x + 4 often appear in equations set to zero. Solving x² + 5x + 4 = 0 gives x = -1 or x = -4. These are the x-intercepts of the parabola y = x² + 5x + 4.
For practice, try factoring 3x² + 5x + 6. The discriminant is 25 - 72 = -47, so no real roots exist. This expression does not factor over the reals. Compare it to x² + 5x + 4, which does.
Understanding how to factor x² + 5x + 4 builds skills needed for solving equations, simplifying rational expressions, and working with polynomials. It's foundational for algebra and calculus.
Mastering factorise x2 5x 4 strengthens algebraic reasoning. It's not just about getting the right answer-it's about seeing the structure behind the symbols.