How do you solve #x^2(a^2+2ab+b^2) = x(a+b)# by factoring?

Question

Nathan Axel · Accepted Answer

#x = 0" " # or #" "x = 1/(a+b)#

First, notice that you can write

#(a^2 + 2ab + b^2)#

as the square of #(a + b)#

#a^2 + 2ab + b^2 = (a + b)^2#

Your equation can now be written as

#x^2 * (a+b)^2 = x * (a + b)#

You can simplify this equation by dividing both sides by #(a+b)#

#(x^2 * (a + b)^color(red)(cancel(color(black)(2))))/color(red)(cancel(color(black)((a+b)))) = (x * color(red)(cancel(color(black)((a+b)))))/color(red)(cancel(color(black)((a+b))))#

#x^2 * (a + b) = x#

Move all the terms to one side of the equation to get

#(a+b) * x^2 - x = 0#

#x * [(a+b)x - 1] = 0#

This equation will have two solutions, #x = color(green)(0)# and

#(a+b)x - 1= 0 implies x = color(green)(1/(a+b))#

Savannah Anderson · Answer

undefined To solve the equation $x^2(a^2+2ab+b^2) = x(a+b)$ by factoring, follow these steps:

1. Factor out $x$ from the right side of the equation:
   $x^2(a^2+2ab+b^2) = x(a+b) $
   $x^2(a^2+2ab+b^2) - x(a+b) = 0 $

2. Factor out $x(a+b)$ from the left side of the equation:
   $x(a+b)(a+b) - x(a+b) = 0 $

3. Notice that both terms have a common factor of $x(a+b)$:
   $x(a+b)((a+b) - 1) = 0 $

4. Simplify:
   $x(a+b)(a+b - 1) = 0 $

5. Apply the zero-product property:
   $x = 0 $ or $a+b = 0 $ or $a+b - 1 = 0 $

6. Solve for $x$:
   $x = 0 $

7. Solve for $a+b$:
   $a+b = 0 $

8. Solve for $a+b - 1$:
   $a+b - 1 = 0 $
   $a+b = 1 $

Therefore, the solutions are $x = 0$, $a+b = 0$, and $a+b = 1$.