Whakaoti i te f(x)=12x^3-17x^2-20x+25

$$f ( x ) = 12 x ^ { 3 } - 17 x ^ { 2 } - 20 x + 25$$

Tauwehe

$$\left(x-1\right)\left(3x-5\right)\left(4x+5\right)$$

Ngā Upane Otinga

Tā te Rational Root Theorem, ko ngā pūtake whakahau katoa o tētahi pūrau kei te āhua o \frac{p}{q}, ina wehea e p te kīanga pūmau 25, ā, ka wehea e q te whakarea arahanga 12. Ko tetahi pūtake pērā ko \frac{5}{3}. Tauwehea te pūrau mā te whakawehe mā te 3x-5.

$$\left(3x-5\right)\left(4x^{2}+x-5\right)$$

Whakaarohia te 4x^{2}+x-5. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4x^{2}+ax+bx-5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=4\left(-5\right)=-20$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -20.

$$-1,20 -2,10 -4,5$$

Tātaihia te tapeke mō ia takirua.

$$-1+20=19 -2+10=8 -4+5=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-4 b=5$$

Tuhia anō te 4x^{2}+x-5 hei \left(4x^{2}-4x\right)+\left(5x-5\right).

$$\left(4x^{2}-4x\right)+\left(5x-5\right)$$

Tauwehea te 4x i te tuatahi me te 5 i te rōpū tuarua.

$$4x\left(x-1\right)+5\left(x-1\right)$$

Whakatauwehea atu te kīanga pātahi x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-1\right)\left(4x+5\right)$$

Me tuhi anō te kīanga whakatauwehe katoa.

$$\left(3x-5\right)\left(x-1\right)\left(4x+5\right)$$

Aromātai

$$\left(x-1\right)\left(3x-5\right)\left(4x+5\right)$$