**Question Corner and Discussion Area**

I need help on how to justify factoring of polynomials. However, I would like an explanation that can be understood by students in pre-Algebra? Can anyone help?I'm not entirely sure what you mean when you want to "justify" factoring of polynomials. Do you mean a mathematical argument that shows why techniques used to factor polynomials are correct and give the right answer, or are you asking what use it is to know how to factor polynomials?

I will assume you are asking the latter, looking for reasons why it is useful to know how to factor polynomials.

The most fundamental reason is to be able to solve polynomial equations. If you have an equation saying that a product of several factors must equal zero, the solution is that one or more of the factors must be zero.

For example, suppose you throw a ball into the air and want to find
when it hits the ground. Laws of physics tell you that the height of
the ball *t* seconds after you threw it is a quadratic polynomial
in *t*. Depending on how fast and at what angle you threw it, you
might, for instance, discover from the laws of physics that the
height of the ball is . To find when it hits the
ground, you want to find what value of *t* gives you . A tough problem as it stands (until you learn about the quadratic
formula). But if you're able to factor the left-hand side, you see that
the equation is saying that and the only time this
product is zero is when one of the factors is zero; in other words, when
*t*=5 or *t*=-1. Ignoring the negative solution (because the ball didn't
start moving until after you threw it), you see that the ball hits the
ground five seconds after you threw it.

You could explain to your students that they will learn a formula for solving any quadratic equation like this, but that factoring is the way such a formula comes about.

Then you could also mention that knowing how a number factors tells you many useful things about it. For example, why is it that if you take any number and take the sum of that number, its cube, and three times its square, you always end up with a multiple of 6? The answer is because the polynomial factors as , and among any three consecutive numbers at least one must be even and at least one must be a multiple of 3, so the product is a multiple of 6.

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