Solving for $x$ from given equation

Question

Sorry for the trivial question. I was wondering if there was an easy way to solve this equation. The equation is:

$$(x-1)(x-2)(x+3)(x-6)+36=0$$

I know this can be solved the traditional way, where after multiplying all terms, we could factorize the entire thing, but I ended up with a really messy equation. Is there a simpler method I'm missing?

Thanks for any help :-)

Answer 1

\begin{align*} (x-1)(x-2)(x+3)(x-6)+36&=0\\ (x^2-3x+2)(x^2-3x-18)+36&=0\\ (x^2-3x)^2-16(x^2-3x)&=0\\ (x^2-3x)(x^2-3x-16)&=0\\ x&=0, 3\textrm{ or }\frac{3\pm\sqrt{73}}{2} \end{align*}

Answer 2

I don't think there's a simpler method of solving the equation without multiplying all the terms and then factorize. Either stick to the traditional way, or use the GDC :)

Answer 3

We have $$ -(x-1)(x-2)(x+3)(x-6)=36, $$ but $36=2^23^2$, so we can easily list the integer solutions, namely $x=0$ and $x=3$.

Answer 4

For more 'complicated' numbers you have to open up the brackets and factorise, however in this case, due to its simple nature, I tried to randomly substitute $x$ by some simple numbers to see if any of them satisfy the equation.

The first two numbers I tried coincidentally worked: $x=0$ and $x=3$. Check for the other two roots yourself, they are probably small integer numbers such as these.

Answer 5

$(x-1)(x-2)(x+3)(x-6)+36$

$=(x-1)(x-6)(x-2)(x+3)+36$

$=(x^2-7x+6)(x^2+x-6)+36$

$=(x^2-3x-4x+6)(x^2-3x+4x-6)+36$

$=(x^2-3x)^2-(4x-6)^2+36$

$=(x^2-3x)^2-((4x-6)^2-6^2)$

$=(x^2-3x)^2-(4x-6+6)(4x-6-6)$

$=(x^2-3x)^2-4x(4x-12)$

$=x^2\times (x-3)^2-16x(x-3)$

$=(x-3)(x^2(x-3)-16x)$

$=(x-3)(x^3-3x^2-16x)$

$=x(x-3)(x^2-3x-16)$

It's a bit hard to find non-integer solutions for $x^2-3x-16$, but I will explain the rest:

• The strategy is to make (part of) the expression equivalent to the equality $(A-B)(A+B)=A^2-B^2$.

• For the lines $3$, $4$, $5$, to identify which is $A$ and which is $B$, we will take the "average" expression and put it $A$, by adding two expressions together and divided by $2$.