$$f(x) = x^2(2x-3)^3$$
I tried to extract the 2 from the parenthesis.
$$f(x) = 2x^2(x-\frac{3}{2})^3$$
But the graphic from this function is different.
What should I consider when doing this kind of factoring?
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$\begingroup$ Don't forget that the terms in the parenthesis are raised to the third power. $\endgroup$– inkievoydNov 10, 2014 at 18:44
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$\begingroup$ @inkievoyd Yes I know something is there, but I don't know what, please if you could explain $\endgroup$– JorgeeFGNov 10, 2014 at 18:45
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1$\begingroup$ $(2x - 3)^3 = (2(x - 3/2))^3 = 8 (x - 3/2)^3$ $\endgroup$– BaronVTNov 10, 2014 at 18:47
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$\begingroup$ You have at least 3 different functions running around now: the original $x^2(2x - 3)^3 (= 8x^2(x -3/2)^3)$, the incorrectly factored $2x^2(x - 3/2)^3$, and a third, different, still incorrect, function in your graphing utility: $x^2(2x - 3/2)^3$ $\endgroup$– BaronVTNov 10, 2014 at 18:53
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2 Answers
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$f(x)=8x^{2}(x-3/2)^{3}$ and not $2x^{2}(x-3/2)^{3}$. This is because since $2(x-3/2)=2x-3$, when taking the third power the $2$ becomes an $8$.
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$\begingroup$ The resulting graphic is even more different than with factor 2 $\endgroup$– JorgeeFGNov 10, 2014 at 18:47
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$\begingroup$ You typed your $f$ function wrong in the graphic. $\endgroup$– SergioNov 10, 2014 at 18:48
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1$\begingroup$ Based on your graphic, it looks like you entered $x^2(2x-\frac32)^3$ and $2x^2(x-\frac32)^3$ neither of which is your original function. $\endgroup$ Nov 10, 2014 at 18:49
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$f(x) = x^2 * ((2)(x-3/2))^3$
= x^2 * (2)^3 (x-3/2)^3
