# 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