Question

Solve the equation $(x + 2)(x + 3)(x + 8)x(x + 12) = 4x^2$.

Answer 1

x = 0, -4, -6

Answer 2

The given equation is $(x + 2)(x + 3)(x + 8)x(x + 12) = 4x^2$.

Step 1: Check for $x=0$.

Substitute $x=0$ into the equation:

$(0 + 2)(0 + 3)(0 + 8)(0)(0 + 12) = 4(0)^2$

$0 = 0$

So, $x=0$ is a solution.

Step 2: Assume $x \neq 0$.

Since $x \neq 0$, we can divide both sides of the equation by $x$:

$(x + 2)(x + 3)(x + 8)(x + 12) = 4x$.

Step 3: Rearrange and group the terms.

We have four terms on the left side: $(x+2)$, $(x+3)$, $(x+8)$, $(x+12)$.

We look for pairs that, when multiplied, result in a common structure, typically $x^2 + Ax + B$.

Let's pair them such that the product of the constants is the same:

Pair 1: $(x+2)(x+12) = x^2 + 12x + 2x + 24 = x^2 + 14x + 24$.

Pair 2: $(x+3)(x+8) = x^2 + 8x + 3x + 24 = x^2 + 11x + 24$.

Substitute these back into the equation:

$(x^2 + 14x + 24)(x^2 + 11x + 24) = 4x$.

Step 4: Introduce a substitution.

Let $A = x^2 + 24$. Then we have $(A + 14x)(A + 11x) = 4x$.

Expanding gives $A^2 + 25Ax + 154x^2 = 4x$.

Substitute $A = x^2 + 24$:

$(x^2 + 24)^2 + 25x(x^2 + 24) + 154x^2 = 4x$

$x^4 + 48x^2 + 576 + 25x^3 + 600x + 154x^2 = 4x$

$x^4 + 25x^3 + 202x^2 + 596x + 576 = 0$

Divide by x, $(x+2)(x+3)(x+8)(x+12) = 4x$

$(x^2+14x+24)(x^2+11x+24)=4x$

$(x^2+14x+24)(x^2+11x+24)-4x=0$

By observation, $x=-4$ is a root:

$((-4)^2+14(-4)+24)((-4)^2+11(-4)+24)-4(-4)=(16-56+24)(16-44+24)+16=(-16)(-4)+16=64+16=80\neq 0$

Let's try to divide the quartic polynomial by $x$.

The equation is $x(x+2)(x+3)(x+8)(x+12) = 4x^2$.

Case 1: $x=0$ is a solution.

Case 2: $x \neq 0$. Then $(x+2)(x+3)(x+8)(x+12) = 4x$.

$(x+2)(x+12) = x^2+14x+24$.

$(x+3)(x+8) = x^2+11x+24$.

$(x^2+14x+24)(x^2+11x+24) = 4x$.

Let $y=x^2+24$, then $(y+14x)(y+11x) = 4x$.

$y^2+25xy+154x^2 = 4x$.

$(x^2+24)^2+25x(x^2+24)+154x^2 = 4x$.

$x^4+48x^2+576+25x^3+600x+154x^2 = 4x$.

$x^4+25x^3+202x^2+596x+576 = 0$.

Since $x \neq 0$, we can divide by $x$:

$x^3+25x^2+202x+596+\frac{576}{x} = 0$.

Let's try $x=-6$:

$(-6+2)(-6+3)(-6+8)(-6+12) = (-4)(-3)(2)(6) = 144$.

$4(-6) = -24$.

Try $x=-4$:

$(-4+2)(-4+3)(-4+8)(-4+12) = (-2)(-1)(4)(8) = 64$.

$4(-4) = -16$.

$x^4+25x^3+202x^2+596x+576 = (x+2)(x+3)(x+8)(x+12)x-4x^2 = 0$.

$x(x+2)(x+3)(x+8)(x+12)-4x^2 = 0$.

$x[(x+2)(x+3)(x+8)(x+12)-4x] = 0$.

$x=0$ is a solution.

$(x+2)(x+3)(x+8)(x+12)-4x = 0$.

$(x+2)(x+12)(x+3)(x+8)-4x = 0$.

$(x^2+14x+24)(x^2+11x+24)-4x = 0$.

$x^4+11x^3+24x^2+14x^3+154x^2+336x+24x^2+264x+576-4x = 0$.

$x^4+25x^3+202x^2+596x+576 = 0$.

$x^4+25x^3+202x^2+596x+576 = (x+4)(x^3+21x^2+118x+144)=0$

$(x+4)(x+6)(x^2+15x+24)=0$

If $x^3 + 21x^2 + 118x + 144 = 0$.

$(x+4)(x^2+17x+36) = 0$.

$(x+4)(x+9)(x+4)=0$

$x = -4, -6, -8$

Therefore $x=0, -4, -6$