How do you factor #w^ { 2} - 32- 4w#?

1 Answer
mimi
Jun 14, 2017

#(w-8)(w+4)#

Explanation:

This equation is in general form, #ax^2+bx+c=0#

You are trying to find two factors that multiply together to be #w^2-32-4w#. It should look like this: #(w+?)(w+?)#

For visual purposes (and to match general form), I rearranged the polynomial:
#w^2-4w-32#

Using this as a reference: #(w+?)(w+?)#
The ?'s should ADD up to equal the b value, which is -4.
The ?'s should be MULTIPLIED to equal the c value, which is -32.

The simplest approach is to find all the FACTORS of -32.
Factors of -32: #(32*-1); (-32*1); (16*-2); (-16*2); (8*-4); (-8*4)#
I put them in pairs to help see the connections between the numbers.

For each pair, add the numbers together until you find a pair with a sum of -4.

You'll see that the pair #(-8*4)# works out to have a sum of -4.

Substitute #-8# and #4# for the ?'s in #(w+?)(w+?)#

The answer is #(w-8)(w+4)#.

It may seem tedious at first, but as you do more problems, you'll get the hang of it and won't have to list numbers out because you can do mental math.

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