How do you solve #x^3 + 147 = 3x^2 + 49x#?
1 Answer
Nov 27, 2015
Subtract
#x = 3# ,#7# or#-7#
Explanation:
Subtract
#x^3-49x = 3x^2-147#
That is:
#x(x^2-49) = 3(x^2-49)#
So either
Alternatively and more systematically:
Subtract the right hand side from the left to get:
#x^3-3x^2-49x+147 = 0#
Let
Factor by grouping:
#x^3-3x^2-49x+147#
#=(x^3-3x^2)-(49x-147)#
#=x^2(x-3)-49(x-3)#
#=(x^2-49)(x-3)#
#=(x^2-7^2)(x-3)#
#=(x-7)(x+7)(x-3)#
...using the difference of squares identity:
#a^2-b^2 = (a-b)(a+b)#
So