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Here is the equation: $$\frac{-3(x+h)^2-6(x+h)+4+3x^2+6x-4}{h}$$

I am getting the wrong answer every time. \begin{align*} f & = \frac{-3(x^2+3xh+h^2)-6x+6h+4+3x^2+6x-4}{h}\\ f & = \frac{-3x^2-6xh-3h^2-6x+6h+4+3x^2+6x-4}{h}\\ f & = \frac{-6xh-3h^2}{h}\\ f & = 3(-2x-h^2) \end{align*}

Two questions, what am I doing wrong when factoring down this problem and where is the best place to really learn factoring super well?

N. F. Taussig
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TGR
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3 Answers3

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Your approach (expanding, collecting terms, and canceling) seems fine but you made four errors. Luckily, two of them cancel each other out: $$-6(x+h) \not = -6x \color{red}{+} 6h \tag{line 1 to 2}$$ $$(x+h)^2 \not = x^2 + \color{red}{3}xh + h^2\tag{line 1 to 2}$$ $$-3(x^2 + 3xh + h^2) \not = -3x^2 -\color{red}{6}xh -3 h^2\tag{line 2 to 3}$$ Finally, you managed to lose a term when you reduced ($6h$, which should be $-6h$).

To improve your factoring:

  • Watch someone else do it. Khan academy has excellent short videos. Your textbook may have examples of some particular method.
  • Practice. Preferably early and often. Preferably in a way that allows non-punitive, immediate feedback. A textbook will have factoring exercises with selected answers.
    • Use pen/pencil & paper.
    • Go slow.
    • Investigate your errors.
    • Try again.
  • Seek out positive reinforcement (from yourself if need be). By the way, nice post! You're already ahead of the curve.
  • Teach someone else. If this is impossible, consider a study group where you might find opportunities to trade teaching roles.
  • Avoid overly competitive or punitive educational environments. If this is impossible, do your learning before you arrive. If you show up at boot camp ready to start your weight loss, you're probably going to wash out of the military. If you do it repeatedly, you'll have zero probability of success.
David Diaz
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it looks like you are differentiating $-(3x^2 + 6x-4)$

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

from which $-3x^2 -6x +4 $ cancel out with $3x^2 +6x -4$

$$ -6xh-3h^2-6h$$

you are missing the $-6h$ term,

and $\frac{-6xh-3h^2-6h}{h}=-6x-3h-6= -3(2x+h+2)$

Furthermore, taking the limit as $h \rightarrow 0$, $-3(2x+2)=-6x-6$.

Check:

$\frac{d}{dx}[-3x^2-6x+4] = (-3)\cdot 2 x -6 +0= -6x-6$.

Included the limit and derivative comments as a bonus, you tagged the question as pre-calculus, I’m not sure what level you’re at, if they confuse you for the moment, you can disregard them.

no lemon no melon
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It is easy to make errors, expanding, collecting terms and cancelling. To reduce typos, I often use WolframAlpha for intermediate results such as here, here, and here.

$$f=\frac{-3(x+h)^2-6(x+h)+4+3x^2+6x-4}{h}\\ =\frac{\big(-3 h^2 - 6 h x - 3 x^2\big)-\big(6 h + 6 x\big)+\big(4+3x^2+6x-4\big)}{h}\\ =\frac{-3 h (h + 2 x + 2)}{h}=-3(h + 2 x + 2)$$

poetasis
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