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)