factoring
- Thread starter Agatha
- Start date
Hello, Agatha!
Many years ago, I "invented" a procedure for this type of problem.
Multiply out: .\(\displaystyle b^2c + bc^2 + ac^2 - a^2c - a^2b - ab^2\)
Arrange the monomials in terms of \(\displaystyle a.\) .**
. . \(\displaystyle \overbrace{-a^2b - a^2c}^{a^2\text{ terms}} \:-\: \overbrace{ab^2 + ac^2}^{a\text{ terms}} \:+\: \overbrace{b^2c + bc^2}^{\text{"constant" terms}}\)
Factor by grouping:
. . \(\displaystyle -a^2(b+c ) - a(b^2-c^2) + bc(b+c)\)
. . \(\displaystyle -a^2(b+c) - a(b-c)(b+c) + bc(b+c)\)
Factor:
. . \(\displaystyle (b+c)\big[-a^2 - a(b-c) + bc\big]\)
. . \(\displaystyle (b + c)\big[-a^2 - ab + ac + bc\big]\)
Factor by grouping:
. . \(\displaystyle (b + c)\big[-a(a + b) + c(a + b)\big]\)
Factor:
. . \(\displaystyle (a+b)(b+c)\big[-a + c\big]\)
. . \(\displaystyle (a+b)(b+c)(c-a)\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** . We can, of course, arrange the monomials in terms of \(\displaystyle b\) or \(\displaystyle c.\)
Many years ago, I "invented" a procedure for this type of problem.
\(\displaystyle \text{Factor: }\:bc(b+c)+ca(c-a)-ab(a+b)\)
\(\displaystyle \text{Answer: }\a+b)(b+c)(c-a)\)
Multiply out: .\(\displaystyle b^2c + bc^2 + ac^2 - a^2c - a^2b - ab^2\)
Arrange the monomials in terms of \(\displaystyle a.\) .**
. . \(\displaystyle \overbrace{-a^2b - a^2c}^{a^2\text{ terms}} \:-\: \overbrace{ab^2 + ac^2}^{a\text{ terms}} \:+\: \overbrace{b^2c + bc^2}^{\text{"constant" terms}}\)
Factor by grouping:
. . \(\displaystyle -a^2(b+c ) - a(b^2-c^2) + bc(b+c)\)
. . \(\displaystyle -a^2(b+c) - a(b-c)(b+c) + bc(b+c)\)
Factor:
. . \(\displaystyle (b+c)\big[-a^2 - a(b-c) + bc\big]\)
. . \(\displaystyle (b + c)\big[-a^2 - ab + ac + bc\big]\)
Factor by grouping:
. . \(\displaystyle (b + c)\big[-a(a + b) + c(a + b)\big]\)
Factor:
. . \(\displaystyle (a+b)(b+c)\big[-a + c\big]\)
. . \(\displaystyle (a+b)(b+c)(c-a)\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** . We can, of course, arrange the monomials in terms of \(\displaystyle b\) or \(\displaystyle c.\)
i have two more questions...first: a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc = ab (a + b + c) + ac (a + b + c) + bc ( a + b + c) = (a + b + c)(ab + ac + bc) ...did i factor this in too few steps? (the solution is correct)
the second...i can't factor this: ab(a - b) - ac(a + c) + bc(2a + c - b)
the solution is: (a - b)(b - c)(a + c)
THANK YOU!!
the second...i can't factor this: ab(a - b) - ac(a + c) + bc(2a + c - b)
the solution is: (a - b)(b - c)(a + c)
THANK YOU!!