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factoring 2
- Thread starter Agatha
- Start date
You could try expanding it out and try grouping appropriate terms.
\(\displaystyle a^{2}b-a^{2}c-ab^{2}+abc+abc-ac^{2}-b^{2}c+bc^{2}\)
Now, factor out a 'b':
\(\displaystyle b(a^{2}-ab+ac-bc)\)
Factor out a 'c':
\(\displaystyle c(a^{2}-ab+ac-bc)\)
See?. What's in the () is the same.
\(\displaystyle b(a^{2}-ab+ac-bc)-c(a^{2}-ab+ac-bc)\)
\(\displaystyle (b-c)(a^{2}-ab+ac-bc)\)
Now, factor what's in the parentheses and finish up
\(\displaystyle a^{2}b-a^{2}c-ab^{2}+abc+abc-ac^{2}-b^{2}c+bc^{2}\)
Now, factor out a 'b':
\(\displaystyle b(a^{2}-ab+ac-bc)\)
Factor out a 'c':
\(\displaystyle c(a^{2}-ab+ac-bc)\)
See?. What's in the () is the same.
\(\displaystyle b(a^{2}-ab+ac-bc)-c(a^{2}-ab+ac-bc)\)
\(\displaystyle (b-c)(a^{2}-ab+ac-bc)\)
Now, factor what's in the parentheses and finish up
D
Deleted member 4993
Guest
can anybody help me with this? ab(a - b) - ac(a + c) + bc(2a + c - b)
i need to factor it :shock:
Did you follow the rule (method) proposed by Soroban in your previous post?!!
http://www.freemathhelp.com/forum/threads/74971-factoring
(bc)[a(a-b)+c(a-b)]=(b-c)(a-b)(a+c)!!!tkank you galactus!!!
sub.khan: obviously i couldn't otherwise you wouldn't see me post on this forum again.as you can see this polinomial is somewhat different,harder because you have to find greatre number of common monomials...and i got lazy since i found out this forum.
big thanks again galactus1
sub.khan: obviously i couldn't otherwise you wouldn't see me post on this forum again.as you can see this polinomial is somewhat different,harder because you have to find greatre number of common monomials...and i got lazy since i found out this forum.
big thanks again galactus1
D
Deleted member 4993
Guest
(bc)[a(a-b)+c(a-b)]=(b-c)(a-b)(a+c)!!!tkank you galactus!!!
sub.khan: obviously i couldn't otherwise you wouldn't see me post on this forum again.as you can see this polinomial is somewhat different,harder because you have to find greatre number of common monomials...and i got lazy since i found out this forum.
big thanks again galactus1
\(\displaystyle a^{2}b-a^{2}c-ab^{2}+abc+abc-ac^{2}-b^{2}c+bc^{2}\)
This is already set up as descending power of 'a' (as Soroban had suggested)
Now Factor out a2 and a (as Soroban had suggested)
\(\displaystyle a^{2}b-a^{2}c-ab^{2}+abc+abc-ac^{2}-b^{2}c+bc^{2}\)
\(\displaystyle = \ a^{2}(b-c) - ab(b-c) + ac(b-c) - bc(b-c)\)
\(\displaystyle = \ (b-c) [a^{2} - ab + ac - cb]\)
\(\displaystyle = \ (b-c) [a(a-b) + c(a - b)]\)
\(\displaystyle = \ (b-c)(a-b)(a+c)\)
It works EXACTLY like the problem before.......
I guess the operative sentence was
i got lazy
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