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Factoring Trinomials with Leading Coefficient 1

Let’s look closely at an example of finding the product of two binomials using the distributive property:

(x + 3)(x + 4) = (x + 3)x + (x + 3)4 Distributive property
  = x2 + 3x + 4x + 12 Distributive property
  = x2 + 7x + 12  

To factor x2 + 7x + 12, we need to reverse these steps. First observe that the coef- ficient 7 is the sum of two numbers that have a product of 12. The only numbers that have a product of 12 and a sum of 7 are 3 and 4. So write 7x as 3x + 4x:

x2 + 7x + 12 = x2 + 3x + 4x+ 12

Now factor the common factor x out of the first two terms and the common factor 4 out of the last two terms. This method is called factoring by grouping.

Rewrite 7x as 3x + 4x.
Factor out common factors.
Factor out the common factor x + 3.

 

Example 1

Factoring a trinomial by grouping

Factor each trinomial by grouping.

a) x2 + 9x + 18

b) x2 - 2x - 24

Solution

a) We need to find two integers with a product of 18 and a sum of 9. For a product of 18 we could use 1 and 18, 2 and 9, or 3 and 6. Only 3 and 6 have a sum of 9. So we replace 9x with 3x  6x and factor by grouping:

x2 + 9x + 18 = x2 + 3x + 6x + 18 Replace 9x by 3x + 6x.
  = (x + 3)x + (x + 3)6 Factor out common factors.
  = (x + 3)(x + 6) Check by using FOIL.

b) We need to find two integers with a product of -24 and a sum of -2. For a product of 24 we have 1 and 24, 2 and 12, 3 and 8, or 4 and 6. To get a product of -24 and a sum of -2, we must use 4 and -6:

x2 - 2x - 24 = x2 - 6x + 4x - 24 Replace -2x with -6x + 4x.
  = (x - 6)x + (x - 6)4 Factor out common factors.
  = (x - 6)(x + 4) Check by using FOIL.

 

The method shown in Example 1 can be shortened greatly. Once we discover that 3 and 6 have a product of 18 and a sum of 9, we can simply write

x2 + 9x + 18 = (x + 3)(x + 6).

Once we discover that 4 and -6 have a product of -24 and a sum of -2, we can simply write

x2 - 2x - 24 = (x - 6)(x + 4).

In the next example we use this shortcut.

 

Example 2

Factoring ax2 + bx + c with a = 1

Factor each quadratic polynomial.

a) x2 + 4x + 3

b) x2 + 3x - 10

c) a2 - 5a + 6

Solution

a) Two integers with a product of 3 and a sum of 4 are 1 and 3:

x2 + 4x + 3 = (x + 1)(x + 3)

Check by using FOIL.

b) Two integers with a product of 10 and a sum of 3 are 5 and -2:

x2 + 3x - 10 = (x + 5)(x - 2)

Check by using FOIL.

c) Two integers with a product of 6 and a sum of -5 are -3 and -2:

a2 - 5a + 6 = (a - 3)(a - 2)

Check by using FOIL.

 

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