The number of solution to the equation $|x|(6x^2 + 1) = 5x^2$ is

Since we have $|x|$, we need to consider:

When $x > 0$, then $|x| = x$

When $x < 0$, then $|x| = -x$

When $x = 0$, then $|x| = 0$

The absolute value $|x|$ behaves differently depending on whether $x$ is positive or negative, so we must handle each situation separately.

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If $x > 0$, then $|x| = x$

$x(6x^2 + 1) = 5x^2$

$6x^3 + x = 5x^2$

$6x^3 - 5x^2 + x = 0$

$x(6x^2 - 5x + 1) = 0$

Now we need to factor $6x^2 - 5x + 1$.

$6x^2 - 5x + 1 = 6x^2 - 2x - 3x + 1$

$= 2x(3x - 1) - 1(3x - 1)$

$= (2x - 1)(3x - 1)$

So our equation becomes:

$x(2x - 1)(3x - 1) = 0$

This gives us: $x = 0$, $x = \dfrac{1}{2}$, or $x = \dfrac{1}{3}$

Since we assumed $x > 0$, we must verify which solutions are valid:

$x = 0$ does NOT satisfy $x > 0$, so we reject this

Valid solutions from Case 1: $x = \dfrac{1}{2}, \dfrac{1}{3}$

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If $x < 0$, then $|x| = -x$

$-x(6x^2 + 1) = 5x^2$

$-6x^3 - x = 5x^2$

$-6x^3 - 5x^2 - x = 0$

$-x(6x^2 + 5x + 1) = 0$

Since we're considering $x < 0$, we have $-x > 0$, so:

$6x^2 + 5x + 1 = 0$

$6x^2 + 5x + 1 = 6x^2 + 2x + 3x + 1$

$= 2x(3x + 1) + 1(3x + 1)$

$= (2x + 1)(3x + 1)$

So: $(2x + 1)(3x + 1) = 0$

This gives us: $x = -\dfrac{1}{2}$ or $x = -\dfrac{1}{3}$

Both values satisfy $x < 0$

Valid solutions from Case 2: $x = -\dfrac{1}{2}, -\dfrac{1}{3}$

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Let's check if $x = 0$ satisfies our original equation:

$|0|(6(0)^2 + 1) = 5(0)^2$

$0 \cdot 1 = 0$

$0 = 0$

Valid solution from Case 3: $x = 0$

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Combining all valid solutions:

From Case 1: $x = \dfrac{1}{2}, \dfrac{1}{3}$

From Case 2: $x = -\dfrac{1}{2}, -\dfrac{1}{3}$

From Case 3: $x = 0$

Total number of solutions: 5