How do you factor #a^2b+6ab^2+9b^2#?

1 Answer
Jan 18, 2017

Answer:

See solution below

Explanation:

To factor this, because all terms are positive and there are two variables we know the factoring will be in the form:

#(color(red)(j)a + color(blue)(k)b)(color(green)(l)a + color(purple)(m)b)#

Where #color(red)(j)#, #color(blue)(k)#, #color(green)(l)# and #color(purple)(m)# are constants.

We also know because of the coefficients in the original problem:
#color(red)(j) xx color(green)(l) = 1#
#color(blue)(k) xx color(purple)(m) = 9#
#(color(red)(j) xx color(purple)(m)) + (color(blue)(k) xx color(green)(l)) = 6#

"Playing" with the factors for #9# (1x9, 3x3, 9x1) Gives:

#(color(red)(1)a + color(blue)(3)b)(color(green)(1)a + color(purple)(3)b)#

#(a + 3b)(a + 3b) = (a + 3b)^2#