How do you factor #3c^3+2c^2-147c-98#?

1 Answer
Jim G.
Oct 8, 2017

#(c-7)(c+7)(3c+2)#

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)#

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