How do you determine whether to use plus or minus signs in the binomial factors of a trinomial of the form x2 + bx + c where b and c may be positive or negative numbers?

Answer 1

Hi Inez V.,

I don't know if you've "been exposed to" the Quadratic Formula yet. But let's assume not. It's a wonderful thing, it immediately solves such factoring problems!

The way it works is this. let's assume an equation of the form ax^2 + bx + c = 0 , where a, b, and c are constants (they may be positive or negative numbers). Then you can factor the trinomial EXACTLY with the following formula: x = (-b ± (b^2 - 4ac)^0.5)/2a . (Note that +/- sign: it stands for "either + or minus" ). The two factors (which are the solutions of the binomials = 0) are then (-b+(b^2-4ac)^0.5)/2a , and (-b-(b^2-4ac)^0.5)/2a . So those binomial expressions that would be the factors are then (x - factor) = 0 so the two binomials would be (x - (-b+(b^2-4ac)^0.5)/2a ) and ( x - (-b-(b^2-4ac)^0.5)/2a ) .

I can almost hear your brain frying as you look at those formulas, but let me show you how they do work out. Let's take an easy example, the original binomials are say (2x - 4) and (x + 5) . Then, the resultant trinomial product is (2x^2 + 6x - 20) , to factor.

All right, what are the a, b, and c in this trinomial? a = 2, b = 6, c = -20 .

Now plug in: (x - (-6+(36+160)^0.5)/4)) and (x - (-6-(36+160)^0.5)/4)) or (x-(-6+14)/4) and (x-(-6-14)/4) or (x-2) and (x+5) . Note that you get back the simplest form of the binomials -- you must look and see that the factor 2 is not directly returned, you must just trial-multiply the two binomials you obtained, to see if there was any such constant multiplied in. Indeed, from the original trinomial, it's not clear how the multiplier of 2 should be approtioned over the tow binomials; there are an infinite number of ways to do that, after all.

Hope this helped you,

-- Cheers, -- Mr. d.

Answer 2

When you are factoring a trinomial, you will get two factors (x +/- a number)(x +/- a number).

First, the signs will follow the rules for multiplication for c: Two positives multiplies will equal a positive (+ * + = +) and two negatives multiplied will equal a positive ( - * - = +). If only one of the numbers in the factors is negative, then c will be negative. That is the first thing to check. Do you need 2 positives, 2 negatives, or 1 positive/1 negative.

Second, the signs and numbers will add to be the sign for b. So you have to think cause and effect a bit.

1) IF c is positive, and b is positive, then both of the factors will be positive.

2) IF c is positive, and b is negative, then both of the factors will be negative.

3) However, if c is negative, then you have to determine which of the factors will be positive and which one will be negative. The biggest factor will take the sign of b because the numbers will subtract to get b.

I know this can be tricky at first, so here are some examples:

1) x² + 7x + 6 = (x + 1)(x + 6, The number factors can't be 2 and 3 - they must be 1 and 6 because they must add to be 7. Both numbers will be positive because c is positive and so is b.

2) x² + 9x + 20 = (x - 4)(x - 5) The number factors can't be 2 and 10 - they must be 4 and 5 because they have to multiply to be 20 and add to be 9. Both numbers will be negative because two negatives will multiply to be +20 for c and add to be -9 for b.

3) x² + x - 6 = (x + 3)(x - 2) The number factors can't be 1 and 6 - they must be 2 and 3 because they will have to multiply to be 6 for c and add to be 1 for b. Since 3 is the larger number, it will need to be + to be able to add +3 and -2 for a +1 for b.

If the signs for 3 and 2 were switched around, you would have this trinomial: x² - x + 6 = (x - 3)(x + 2).

In order to factor trinomials easily, you must have a firm grasp of signs in adding/subtracting integers, signs in multiplying integers, and switch between those concepts without becoming confused. It isn't hard, but it does take lots of practice!

Best wishes,

Sue