Random Quadratic Equation Generator

The quadratic ax² + bx + c in factored form is (dx + e)(fx + g). For example, 4x² + 5x - 6 (with a = 4, b = 5, c = -6) in factored form is (4x-3)(x+2) (with d = 4, e = -3, f = 1, g = 2). The goal of this website is for you to practice factoring, in other words finding d, e, f, and g. The website generates random quadratics but you get to set the difficulty.

Problem Solving

Factoring a quadratic ax² + bx + c is the reverse problem of expanding (dx + e)(fx + g) like division is the reverse problem of multiplication. Just as knowing your times tables helps a student perform effective division, understanding expansion of (dx + e)(fx + g) aids in being proficient at factoring. We know that ax² + bx + c = (dx + e)(fx + g). To find d, e, f, and g, it is illuminating to carry out an expansion.

ax2 + bx + c	=	(dx + e)(fx + g)
	=	dx(fx + g) + e(fx + g)
	=	dfx2 + dgx + efx + eg
	=	(df)x2 + (dg + ef)x + (eg)

Or more visually:

*	dx	e
fx	dfx2	efx
g	dgx	eg

From the expansion, by matching like terms we see that a = df, b = dg + ef, and c = eg. This is useful since d and f must be factors of a, and e and g must be factors of c. Now, the problem reduces to finding the right set of factors. If we consider the example 4x² + 5x - 6 with a = 4 and c = -6, then the factors of 4 are 1, 2, and 4, and the factors of -6 are -1, 1, -2, 2, -3, 3, -6. Then the possible choices for d and f are (1, 4) and (2, 2) and the possible choices for e and g are (-1, 6), (1, -6), (-2, 3), (2, -3). By trying out these combinations, we find that d = 4, e = -3, f = 1, g = 2 or (4x - 3)(x + 2) = 4x² + 5x - 6. There are a lot of different combinations since a = 6 and c = 4 have multiple factors. Then by looking at the number of factors coefficients a and c have, we can gauge the difficulty of the problem.

Instructions

Set the minimum and maximum values for d, e, f, and g. Then click the bottom button to generate equation.

Settings

The difficulty of the problem dependent on the number of possible factors. Let n be the number of factors of a and m be the number of factors of c. Then the total number of choices to consider is proportional to m*n.

• Easy: a = 1 so set d = f = 1. Put the following (min val d = 1, max val d = 1, min val f = 1, max val f = 1).
• Medium: set a to be a small prime number less than or equal to 10 (or other small number by replacing 10). Put in (min val d = 1, max val d = 10, min val f = 1, max val f = 10, and check off "Coefficient a is prime" box).
• Harder: use default settings which can be done by refreshing page.

Min value for d: Max value for d:

Min value for e: Max value for e:

Min value for f: Max value for f:

Min value for g: Max value for g:

Generate Random Equation Hide/Show Solutions

Random Rational Equation

Generate Rational Equation Hide/Show Solutions

Equation with Radical

To clarify, nn the two solutions listed below, the first one is extraneous while the second is not extraneous. $$\text{2x + 3}\sqrt{\text{x+5}}\text{= 11}$$ Generate Radical Equation Hide/Show Solutions