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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 )=nn+n3+2n+23 = n 2 +3n+2n+6 = n 2 +5n+6

But how can you start with the answer and find the factors?

Factoring x 2 +bx+c when b and c are both positive

Example 1:

Factor n 2 +8n+15 .

The factoring can be done by finding two numbers whose sum is 8 and product is 15 :

List pairs of numbers that have a product of 15 and look for a pair that adds to 8 .

( 1,15 ) 1×15=15 1+15=16 ( 3,5 ) 3×5=15 3+5=8

So, using 3 and 5 we can re-write the expression this way:

n 2 +8n+15

= n 2 +3n+5n+15 . . . . by re-writing 8n as 3n+5n

=( n 2 +3n )+( 5n+15 ) …split the expression into two parts

=n( n+3 )+5( n+3 ) . . . .factor each part using the Distributive Property

=( n+5 )( n+3 ) . . . . . . . use the Distributive Property again to extract the factor ( n+3 )

So, the factored form of n 2 +8n+15 is =( n+5 )( n+3 ) .

Example 2:

Factor x 2 +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 x 2 +37x+100 cannot be factored (that is, it is an irreducible polynomial).

But do you see how you would factor x 2 +29x+100 ?

Factoring x 2 +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 x 2 7x+10 .

The negative factor pairs for 10 are:

10=( 10 )( 1 );101=11 10=( 5 )( 2 );52=7

So the polynomial can be factored as

x 2 7x+10=( x2 )( x5 ) .

Factoring x 2 +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 x 2 +6x16 .

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

x 2 +6x16=( x2 )( x+8 ) .

Example 5:

Factor x 2 x20 .

Here we need to find two numbers with opposite signs which have 20 as a product and 1 as a sum.

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

x 2 x20=( x5 )( x+4 ) .

Factoring a x 2 +bx+c,a1

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 14 x 2 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.

            14 x 2 2x35x+5
Factor common factors in pairs and use the Distributive Property.

           ( 14 x 2 2x )( 35x5 )

           2x( 7x1 )5( 7x1 )
                                   
Again, use the Distributive Property.                                         

           ( 7x1 )( 2x5 )

See also factoring by grouping and irreducible polynomials.
 
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