Now you just need to put the pieces together to find the values of m and n. You know that x is the first term of both binomials, and you know that the sum of m and n is 10 and the product of m and n is 21. The pair of numbers that fit the bill for m and n are 3 and 7. Thus, x2 + 10x + 21 = (x + 3)(x + 7). The quadratic expression has now been factored and simplified.
On the Math IC, though, you will often be presented with a quadratic equation. The only difference between a quadratic equation and a quadratic expression is that the equation is set equal to 0 (x2 + 10x + 21 = 0). If you have such an equation, then once you have factored the quadratic you can solve it. Because the product of two terms is zero, one of the terms must be equal to zero. Thus, since x + 3 = 0 or x + 7 = 0, the solutions (also known as the roots) of the quadratic must be x = –3 and x = –7.
Quadratics with Negative Terms
So far we’ve dealt only with quadratics in which the terms are all positive. Factoring a quadratic that has negative terms is no more difficult, but it might take slightly longer to get the hang of it, simply because you are less used to thinking about negative numbers.
Consider the quadratic equation x2 – 4x – 21 = 0. There are a number of things you can tell from this equation: the first term of each binomial is x, since the first term of the quadratic is x2; the product of m and n is –21; and the sum of a and b equals –4. The equation also tells you that either m or n must be negative but that both cannot be negative, because the multiplication of one positive and one negative number can only result in a negative number. Now you need to look for the numbers that fit these requirements for m and n. The numbers that multiply together to give you –21 are: –21 and 1, –7 and 3, 3 and –7, and 21 and –1. The pair that works in the equation is –7 and 3.
Two Special Quadratic Polynomials
There are two special quadratic polynomials that pop up quite frequently on the Math IC, and you should memorize them. They are the perfect square and the difference of two squares. If you memorize the formulas below, you may be able to avoid the time taken by factoring.
There are two kinds of perfect square quadratics. They are:
- a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2. Example: a2 + 6ab + 9 = (a + 3)2
- a2 – 2ab + b2 = (a – b)(a – b) = (a – b)2. Example: a2 – 6ab + 9 = (a –3)2
Note that when you solve for the roots of a perfect square quadratic equation, the solution for the equation (a + b)2 = 0 will be –b, while the solution for (a + b)2 = 0 will be b.
The difference of two squares quadratics follow the form below:
Here’s an instance where knowing the perfect square or difference of two square equations can help you:
