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Trial-and-Error Method

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Posted by dolphin@eskimo.com (64.24.215.27) on November 18, 2001 at 17:35:31:

In Reply to: Algebra - Factoring Trinomials & Checking with FOIL posted by on November 18, 2001 at 14:27:42:

There are two ways to approach this problem. One way uses trial-and-error, and the other way uses factoring by grouping. I'll describe both ways, and you can choose which way you prefer.

With the trial-and-error method, we start by assuming that the trinomial factors as a product of two binomials. In other words, we assume that the factors take the following form:

(a*r + b)(c*r + d)

where we have to find a, b, c, and d.

Well, the first term in the trinomial is 25r^2, and it comes from multiplying a*r and c*r together (the "firsts" in FOIL). So we know that a and c must be factors of 25. Now we ask ourselves, what are the factors of 25?

The factors of 25 are:

(1)(25)
(-1)(-25)
(5)(5)
(-5)(-5)

So the factors of our trinomial must take one of the following forms:

(r + b)(25r + d)
(-r + b)(-25r + d)
(5r + b)(5r + d)
(-5r + b)(-5r + d)

Now, the last term in our trinomial is 9. It comes from multiplying b*d (the "lasts" in FOIL). This means that b and d must be factors of 9.

The factors of 9 are:

(1)(9)
(-1)(-9)
(3)(3)
(-3)(-3)

So, our factoring has to be one of the following:

(r + 1)(25r + 9)
(r + 9)(25r + 1)
(r - 1)(25r - 9)
(r - 9)(25r - 1)
(r + 3)(25r + 3)
(r - 3)(25r - 3)

(-r + 1)(-25r + 9)
(-r + 9)(-25r + 1)
(-r - 1)(-25r - 9)
(-r - 9)(-25r - 1)
(-r + 3)(-25r + 3)
(-r - 3)(-25r - 3)

(5r + 1)(5r + 9)
(5r - 1)(5r - 9)
(5r + 3)(5r + 3)
(5r - 3)(5r - 3)

(-5r + 1)(-5r + 9)
(-5r - 1)(-5r - 9)
(-5r + 3)(-5r + 3)
(-5r - 3)(-5r - 3)

The last step is checking these possibilities using FOIL until we find the right one.

My other post describes factoring by grouping.

~ Mark