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Explain how to factor the following trinomials forms: x² + bx + c

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3 Answers

For x2+bx+c, notice that (x - r1)(x - r2) = x2 - (r1+r2)x + r1r2.

Thus we want integers r1 and r2 with a sum -b and product c.

For example, x2 + 7x + 10.

Notice that the two integers with a sum -7 and a product of 10 are -2 and -5

Thus we get (x+2)(x+5)

For the more general case ax2+bx+c, if it factors then we can write b = b1 + b2 where the following proportion holds:

a:b1 = b2:c

or equivalently, b1b2 = ac

For example, 2x2 + 11x + 12.

We want b1 + b2 = 11 and  b1b2 = 2*12 = 24

We can take b1 = 3 and b2 = 8 and get

2x2 + 11x + 12 = 2x2 + 3x + 8x + 12 = x(2x + 3) + 4(2x + 3) = (x + 4)(2x + 3)

 

There is formula "ax2 + bx + c = a(x-x1)(x-x2) , x1 and x2 are roots of equation ax2 + bx + c = 0 "
x1,2 = [-b ± √(b2 - 4ac)] / 2a

2x2 + 5x + 3 = 0 , x1,2 = [-5 ± √(25 - 24)] / 4 ,
x1 = (-5 + 1)/4 = -1
x2 = (-5 - 1)/4 = -6/4 = -3/2
2(x-(-1))(x-(-3/2)) = (x+1)(2x+3)

When you have ax2 + bx + c, the quickest way to factor it is to first multiply a by c, and then find two factors that will multiply to equal a*c , add up to b. Then you write (ax + _)(ax + _) with the two factors in the blank spots, then divide it all by a.

Eg. 5x2 + 8x + 3 

Here, a*c = 5*3 = 15 and b = 8

Factors that multiply to 15 and add up to 8, are 5 and 3, since 5+3 = 8.

Then write it like so, and you can pull a factor of 5 from the first set of parentheses:

(5x+5)(5x+3) = 5(x+1)(5x+3) but remember we multiplied a by c earlier, so we've actually multiplied the original expression by a (and a = 5), so we need to divide it by a, which we can do to get:

(x+1)(5x+3)

 

Another example:

2x² + 5x + 3

a*c = 2*3 = 6 and b = 5

Factors that add up to 5 and multiply to 6, are: 2 and 3.

So write it as:

(2x + 3)(2x+2) and now divide by a (which is 2): 

(2x + 3)(x+1)

 

Another example

5x2 - 16x + 3

a*c = 5*3 = 15 and b = -16

the factors are... (-1)*(-15) = 15 and (-1) + (-15) = -16

(5x - 15)(5x - 1) now divide by a (which is 5)

(x - 3)(5x - 1)

Comments

Hi Daniel O.I’m a little confused. Are you saying to divide the second set of parenthesis by 2? Because if I divide (2x + 3)(2x+2) by 2 that is totally different than just dividing the second set of equations. Does this work with all trinomials? Where did you get the “2” that you are dividing?

You can divide either set of parentheses by (that's where the 2 came from, see the other examples where i divide by a), but for convenience, divide the one that'll give you whole numbers, not fractions. We divide by a because we multiplied by a when finding a*c.

And yes, this method works for all trinomials.

"if I divide (2x + 3)(2x+2) by 2 that is totally different than just dividing the second set of equations."

It won't matter where you divide, as long as you divide (2x + 3)(2x+2) by 2. You could divide the first parentheses by 2 to get:

(x + 1.5)(2x+2) which is equal to (2x + 3)(x+1) - try FOILing them both.

This is just one of the properties of division. If you have (a*b)/2, it is equal to (a/2)*b and a(b/2).

Comment