How do you factor 3c^3+2c^2-147c-983c3+2c2−147c−98?
1 Answer
Oct 8, 2017
Explanation:
"rearranging the terms "rearranging the terms
(3c^3-147c)+(2c^2-98)(3c3−147c)+(2c2−98)
"factorising each pair"factorising each pair
=color(red)(3c)(c^2-49)color(red)(+2)(c^2-49)=3c(c2−49)+2(c2−49)
"take out the "color(blue)"common factor "(c^2-49)take out the common factor (c2−49)
=(c^2-49)(color(red)(3c+2))to(color(red)(1))=(c2−49)(3c+2)→(1)
c^2-49" is a "color(blue)"difference of squares"c2−49 is a difference of squares
•color(white)(x)a^2-b^2=(a-b)(a+b)∙xa2−b2=(a−b)(a+b)
"here "a=c" and "b=7here a=c and b=7
rArrc^2-49=(c-7)(c+7)⇒c2−49=(c−7)(c+7)
"returning to " (color(red)(1))returning to (1)
rArr(c^2-49)(3c+2)=(c-7)(c+7)(3c+2)⇒(c2−49)(3c+2)=(c−7)(c+7)(3c+2)
