SOLUTION: 3.) 4x2 + 20x + 5y +xy 4.) 48x2y + 20xz +12xy + 5z

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Question 122586: 3.) 4x2 + 20x + 5y +xy
4.) 48x2y + 20xz +12xy + 5z

Found 2 solutions by edjones, jim_thompson5910:
Answer by edjones(7311) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2 + 20x + 5y +xy
=4x^2+20x+xy+5y
=4x(x+5)+y(x+5)
=(4x+y)(x+5)
.
48x^2y + 20xz +12xy + 5z
48x^2y+12xy+20xz+5z
12xy(4x+1)+5x(4x+1)
(12xy+5x)(4x+1)
.
Ed

Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
# 3

4x%5E2+%2B+20x+%2B+5y+%2Bxy Start with the given expression

4x%5E2%2B20x%2Bxy%2B5y Switch the last two terms

%284x%5E2%2B20x%29%2B%28xy%2B5y%29 Group like terms


4x%28x%2B5%29%2By%28x%2B5%29 Factor out the GCF 4x out of the first group. Factor out the GCF y out of the second group


%284x%2By%29%28x%2B5%29 Since we have the common term x%2B5, we can combine like terms

So 4x%5E2+%2B+20x+%2B+5y+%2Bxy factors to %284x%2By%29%28x%2B5%29







# 4



48x%5E2y%2B20xz%2B12xy%2B5z Start with the given expression

%2848x%5E2y%2B20xz%29%2B%2812xy%2B5z%29 Group like terms


4x%2812xy%2B5z%29%2B1%2812xy%2B5z%29 Factor out the GCF 4x out of the first group. Factor out the GCF 1 out of the second group


%284x%2B1%29%2812xy%2B5z%29 Since we have the common term 12xy%2B5z, we can combine like terms

So 48x%5E2y%2B20xz%2B12xy%2B5z factors to %284x%2B1%29%2812xy%2B5z%29