factoring trinomials

Question

How do I factor a trinomial where the coefficient is greater than one? why are they more difficult?

Answer 1

Here's a good way to factor trinomials with leading coefficients that you can't factor out:

 

 

 

1. For a trinomial ax² + bx + c, find a pair of numbers that multiply together to be a×c and add together to be b. Call that pair of numbers p and q.

 

2. Replace bx with px + qx.

 

3. Factor by grouping.

 

 

 

Here's an example of how that works:

 

y = 3x² - 5x - 2

 

 

 

1. Find a pair of numbers that multiply together to be a×c and add together to be b

 

We're trying to find a pair of numbers that multiplies together to be a×c = 3×(-2) = -6 and adds together to be -5. This step is a lot like factoring trinomials without leading coefficients.

 

Factor pairs of -6 are:

 

1×-6

-1×6

2×-3

-2×3

 

Those add up to:

 

1+-6 = -5

-1+6 = 5

2+-3 = -1

-2+3 = 1

 

So, 1 and -6 multiply together to be -6 and add together to be -5.

 

 

 

2. Replace bx with px + qx

 

y = 3x² - 5x - 2

y = 3x² + 1x - 6x - 2

y = 3x² + x - 6x - 2

 

 

 

3. Factor by grouping

 

y = (3x² + x) + (-6x - 2)

y = x(3x + 1) - 2(3x + 1)

y = (x - 2)(3x + 1)

 

 

 

This is the standard method for factoring trinomials with leading coefficients not equal to 1. It's not too bad, and works a lot like the method for factoring simple trinomials.

 

 

 

And, if you can't find a factor pair that works, you can always use the Quadratic Formula, like Nathan says. Be careful, though... For my example, you'll get 2 and -1/3 as your answers, but it's not correct to write:

 

y = (x - 2)(x + 1/3)

 

Instead, always multiply the whole thing by the leading coefficient to get:

 

y = 3(x - 2)(x + 1/3)

 

That's equivalent to the answer we got above:

 

y = (x - 2)3(x + 1/3)

y = (x - 2)(3x + 1)

Answer 2

The first part is to see if there is a common factor in the coefficients.

Example:

2x² + 10x + 12

Here, you can see that there is a 2 in common with all of the parts, so we can factor that out:

2(x² + 5x + 6)

 

From here it's just a matter of standard factorization:

2 + 3 = 5

2 * 3 = 6

2(x + 2)(x + 3) is the final factorization.

 

The reason that they are much more difficult is that there are a greater number of possible factorizations.

Example:

2x² + 15x + 18

Here we don't have anything easily factorable. We know that there is going to be a 2x and an x (and both positive due to the signs in front of the 15 and 18:

(2x + )(x + )

 

18 factors all the way to 2 * 3 * 3. (Another possibility would be 18 * 1, but due to 15 being smaller yet not near 0, we can thankfully ignore that option.) Is it factored to 2*3 and 3, or 2 and 3*3? Not only that, but which one goes with the 2x set, and which one belongs to the x set? This admittedly can take a bit of trial and error due to now having:

M * N = 18

2M + N = 15 or M + 2N = 15

 

15 is close to 18, but smaller, so the 3*3 with the x set isn't going to happen (we would get (2x + 2)(x + 9), giving us too much x value, but rather close). Let's try something a little smaller, then. with 2*3 with the x set:

(2x + 3)(x + 6)

2x² + 12x + 3x + 18 --> 2x² + 15x + 18

 

If you want a tool to help you supersede the trial and error, there's the quadratic formula:

in the equation ax² +bx + c, the formula reads as:

(-b ± √(b² - 4ac)) / 2a when y = 0

 

If we use my equation, we would have:

x = (-15 ± √(15² - 4 * 2 * 18)) / 2*2

x = (-15 ± √(225 - 144)) / 4

x = (-15 ± √81) / 4

x = (-15 ± 9) / 4

x =  -24/4 or 6/4

x = -6 or -3/2

-6 + what = 0

and

-3/2 + what = 0?

The answers are 6 and 3/2.

Since one of them has a 2 denominator, that must be part of the 2x set, and the 6 as a whole must be part of the x set.