Factoring
You can use the distributive law to see that
3(4n + 5) = 12n + 15,
and you can use FOIL to see that
(n + 2)(n + 3) = n2 + 5n + 6.
But how can you start with the answer and find the factors?
Factoring x2 + bx + c when b and c are both positive
Example 1:
Factor n2 + 8n + 15.
This can be done (factored) by finding two numbers whose sum is 8 and product is 15: use 3 and 5. So we get:
n2 + 8n + 15
= n2 + 3n + 5n + 15 . . . . now factor common factors in pairs
= n(n + 3) + 5(n + 3) . . . .use the Distributive Property
= (n + 5)(n + 3) . . . . . . . the (n + 3) was the common factor!
Example 2:
Factor x2 + 37x + 100.
We need two numbers whose product is 100 and sum is 37.
100 = (100)(1); 100 + 1 = 101
100 = (50)(2) ; 50 + 2 = 52
100 = (25)(4) ; 25 + 4 = 29
100 = (20)(5) ; 20 + 5 = 25
100 = (10)(10); 10 + 10 = 20.
It seems that 37 never came up as a sum, so x2 + 37x + 100 cannot be factored (that is, it is an irreducible polynomial).
But do you see how you would factor x2 + 29x + 100?
Factoring x2 + bx + c when b is negative, c is positive
In this case, you need two negative numbers, so that their product is positive but their sum is negative.
Example 3:
Factor x2 – 7x + 10.
The negative factor pairs for 10 are:
10 = (–10)(–1) ; –10 – 1 = –11
10 = (–5)(–2) ; –5 – 2 = –7
So the polynomial can be factored as
x2 – 7x + 10 = (x – 2)(x – 5).
Factoring x2 + bx + c when c is negative
In this case, you need two numbers with opposite signs (so that their product is negative).
Example 4:
Factor x2 + 6x – 16.
Here we need to find two numbers with opposite signs which have –16 as a product and 6 as a sum.
The factor pairs for –16 are:
–16 = (–16)(1) ; –16 + 1 = –15
–16 = (–8)(2) ; –8 + 2 = –6
–16 = (–4)(4) ; –4 + 4 = 0
–16 = (–2)(8) ; –2 + 8 = 6
–16 = (–1)(16); –1 + 16 = 15
–2 and 8 work. So we can factor the polynomial as
x2 + 6x – 16 = (x – 2)(x + 8).
Example 5:
Factor x2 – x – 20.
Here we need to find two numbers with opposite signs which have –20 as a product.
The factor pairs for –20 are:
–20 = (–20)(1) ; –20 + 1 = –19
–20 = (–10)(2) ; –10 + 2 = –8
–20 = (–5)(4) ; –5 + 4 = –1
–20 = (–4)(5) ; –4 + 5 = 1
–20 = (–2)(10) ; –2 + 10 = 8
–20 = (–1)(20) ; –1 + 20 = 19
–5 and 4 work. So we can factor the polynomial as
x2 – x – 20 = (x – 5)(x + 4).
Factoring ax2 + bx + c, a ≠ 1
Things get a little trickier in this case. We need to find two numbers whose product is equal to the product of the leading coefficient and the constant and whose sum is equal to the coefficient of the x term.
Example:
Factor 14x2 – 37x + 5.
Multiply the leading coefficient by the constant
(14)(5) = 70
Find the factor pairs that multiply to 70 and add to –37.
–2 and –35
Replace the middle term.
14x2 – 2x – 35x + 5
Factor common factors in pairs and use the Distributive Property.
(14x2 – 2x) – (35x – 5 )
2x(7x – 1) – 5(7x – 1 )
Again, use the Distributive Property.
(7x – 1)(2x – 5)
