How do you calculate the reverse FOIL method step-by-step?
First, identify a, b, and c in ax² + bx + c. Then find two numbers p and q such that p × q = a·c and p + q = b. Rewrite the middle term bx as px + qx, then factor by grouping to get the two binomial factors. Finally, divide by the leading coefficient a if needed.
How do you factorize x² + 4x + 3?
Here a=1, b=4, c=3, so a·c = 3. Find two numbers that multiply to 3 and add to 4: those are 1 and 3. Rewrite as x² + 1x + 3x + 3, group as x(x+1) + 3(x+1), and factor to get (x+1)(x+3).
How do you solve a quadratic equation using factorization?
Once the trinomial is factored into (x + r)(x + s) = 0, set each factor equal to zero: x + r = 0 gives x = −r, and x + s = 0 gives x = −s. These two values are the roots (solutions) of the quadratic equation. You might also find our Line Equation from Two Points Calculator useful.
What if the discriminant is negative?
If b² − 4ac < 0, the trinomial has no real factor pairs and cannot be factored over the real numbers using the reverse FOIL method. The roots in that case are complex (imaginary) numbers.
Can the reverse FOIL method be used when a ≠ 1?
Yes. When the leading coefficient a is not 1, you multiply a × c to get the target product, find two numbers that multiply to a·c and add to b, split the middle term, factor by grouping, and then simplify. This is sometimes called the AC method or splitting the middle term.
What is the difference between FOIL and reverse FOIL?
FOIL (First, Outer, Inner, Last) is the process of multiplying two binomials to produce a trinomial. Reverse FOIL works backwards — it takes a trinomial and finds the two binomials whose product equals that trinomial. They are inverse operations of each other.
When is a trinomial not factorable using reverse FOIL?
A trinomial ax² + bx + c is not factorable over the integers when no integer pair (p, q) exists such that p × q = a·c and p + q = b. This happens when the discriminant (b² − 4ac) is not a perfect square. The trinomial is then called 'prime' or 'irreducible' over the integers.