How do you factor #3c^3+2c^2-147c-98#?
1 Answer
Oct 8, 2017
Explanation:
#"rearranging the terms "#
#(3c^3-147c)+(2c^2-98)#
#"factorising each pair"#
#=color(red)(3c)(c^2-49)color(red)(+2)(c^2-49)#
#"take out the "color(blue)"common factor "(c^2-49)#
#=(c^2-49)(color(red)(3c+2))to(color(red)(1))#
#c^2-49" is a "color(blue)"difference of squares"#
#•color(white)(x)a^2-b^2=(a-b)(a+b)#
#"here "a=c" and "b=7#
#rArrc^2-49=(c-7)(c+7)#
#"returning to " (color(red)(1))#
#rArr(c^2-49)(3c+2)=(c-7)(c+7)(3c+2)#