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

1 Answer
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)