Explain how

to factor the following trinomials forms: x² + bx + c (Example: x² + 4x + 4) and ax² + bx + c (Example: 2x² + 5x + 3). Be specific and show your answer using both

words and mathematical notation.

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

# 3 Answers

For x^{2}+bx+c, notice that (x - r_{1})(x - r_{2}) = x^{2} - (r_{1}+r_{2})x + r_{1}r_{2}.

Thus we want integers r_{1} and r_{2} with a sum -b and product c.

For example, x^{2} + 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 = b_{1} + b_{2} where the following proportion holds:

a:b_{1} = b_{2}:c

or equivalently, b_{1}b_{2} = ac

For example, 2x^{2 }+ 11x + 12.

We want b_{1} + b_{2} = 11 and b_{1}b_{2} = 2*12 = 24

We can take b_{1} = 3 and b_{2} = 8 and get

2x^{2} + 11x + 12 = 2x^{2} + 3x + 8x + 12 = x(2x + 3) + 4(2x + 3) = (x + 4)(2x + 3)

There is formula "**ax ^{2} + bx + c = a(x-x_{1})(x-x_{2})** , x

_{1}and x

_{2}are roots of equation ax

^{2}+ bx + c = 0 "

x

_{1,2}= [-b ± √(b

^{2}- 4ac)] / 2a

2x

^{2}+ 5x + 3 = 0 , x

_{1,2 }= [-5 ± √(25 - 24)] / 4 ,

x

_{1 }= (-5 + 1)/4 = -1

x

_{2 }= (-5 - 1)/4 = -6/4 = -3/2

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

**(x+1)(2x+3)**

When you have **ax ^{2} + 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. 5x^{2} + 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

5x^{2} - 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

You can divide either set of parentheses by

a(that's where the 2 came from, see the other examples where i divide bya), but for convenience, divide the one that'll give you whole numbers, not fractions. We divide byabecause we multiplied byawhen findinga*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