Differentiate h(x) = (2x + 1)⁵(x² − 1)³ and factor the result.

Step-by-step solution

1. 1

   Set up the product rule. The function is a product of two composite powers, so write

   $u = (2x+1)^5, \qquad v = \left(x^2-1\right)^3, \qquad h' = u'v + uv'$

   Expanding $h$ first would give a degree-11 polynomial; the product rule keeps everything in factored form, which is what makes the final simplification possible.

2. 2

   Differentiate each factor with the chain rule. Outer derivative times inner derivative:

   $u' = 5(2x+1)^4 \cdot 2 = 10(2x+1)^4$

   $v' = 3\left(x^2-1\right)^2 \cdot 2x = 6x\left(x^2-1\right)^2$

   The inner derivatives $2$ and $2x$ are exactly where a chain-rule slip usually happens.

3. 3

   Assemble the two terms.

   $h'(x) = 10(2x+1)^4\left(x^2-1\right)^3 + 6x(2x+1)^5\left(x^2-1\right)^2$

   Each term differs from $h$ by dropping one power from one factor, which is why a common factor is guaranteed to exist.

4. 4

   Factor out the lowest power of each bracket. Both terms contain $(2x+1)^4$ and $\left(x^2-1\right)^2$:

   $h'(x) = (2x+1)^4\left(x^2-1\right)^2\Big[10\left(x^2-1\right) + 6x(2x+1)\Big]$

5. 5

   Simplify the bracket and pull out the numerical factor.

   $10x^2 - 10 + 12x^2 + 6x = 22x^2 + 6x - 10 = 2\left(11x^2 + 3x - 5\right)$

   $h'(x) = 2(2x+1)^4\left(x^2-1\right)^2\left(11x^2+3x-5\right)$

   The last quadratic has discriminant $9 + 220 = 229$, not a perfect square, so it does not factor further over the rationals.

6. 6

   Check the result numerically and read off the critical points. A central difference at $x = 0$ gives $h'(0) \approx -10.000$, matching $2(1)(1)(-5) = -10$; the same check at $x = 0.3,\ 1.7,\ -2.1$ agrees to six significant figures. The zeros of $h'$ are $x = -\tfrac12$, $x = \pm 1$ (from the repeated factors, which are also zeros of $h$) and $x = \frac{-3 \pm \sqrt{229}}{22}$.