# factoring and solving $(x^2-4x+4)(x^2+6x+9)=2x^2+2x+12$

I need to solve this equation $$(x^2-4x+4)(x^2+6x+9)=2x^2+2x+12$$

If I simplify I'll get $4$th degree.

Is there any simple method to factorize and solve?

I can see that: $$(x-2)^2\cdot(x+3)^2=2(x^2+x+6)$$

I'm not allowed to use polynomial division.

Wolfram said that it's equal to: $$(x^2+x-6)^2=2(x^2+x+6)$$ and then it's easy because I can put $t=x^2+x$ but I don't know how to get this factorization.

Thanks.

From $(x-2)^2\cdot(x+3)^2=2(x^2+x+6)$ you do $(x-2)(x+3)=x^2+x-6$ twice on the left to get what you want. Without knowing the answer I am not sure I would see that.

• thanks i don't think i would see that even with wolfram.
– dani
Jan 8 '15 at 23:53

To me, this question is more on the side of pattern recognition rather than factoring.
Notice in the equation $$(x^2-4x+4)(x^2+6x+9)=2x^2+2x+12$$ The RHS is very close to the sum of the two factors on LHS. In fact, $$RHS = (x^2-4x+4) + (x^2 + 6x + 9) - 1$$ This leads to \begin{align} & (x^2-4x+4)(x^2+6x+9) = (x^2-4x+4) + (x^2 + 6x + 9) - 1\\ \iff & ((x^2-4x+4)-1)((x^2+6x+9)-1) = 0\\ \iff & (x^2-4x+3)(x^2+6x+8) = 0\\ \iff & (x-1)(x-3)(x+2)(x+4) = 0\\ \implies & x = 1, 3, -2 \text{ or } -4. \end{align}

Hint: $$(x-2)(x+3) = x^2+ x - 6$$

It's actually very simple

$$(x-2)^2(x+3)^2 = \big[(x-2)(x+3)\big]^2 = (x^2+x-6)^2$$