Whakaoti i te =6x^3+13x^2-13x-20

$$= 6 x ^ { 3 } + 13 x ^ { 2 } - 13 x - 20$$

Tauwehe

$$\left(3x-4\right)\left(x+1\right)\left(2x+5\right)$$

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 -20, ā, ka wehea e q te whakarea arahanga 6. Ko tetahi pūtake pērā ko -\frac{5}{2}. Tauwehea te pūrau mā te whakawehe mā te 2x+5.

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

Whakaarohia te 3x^{2}-x-4. Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 3x^{2}+ax+bx-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=3\left(-4\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

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

$$a=-4 b=3$$

Tuhia anō te 3x^{2}-x-4 hei \left(3x^{2}-4x\right)+\left(3x-4\right).

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

Whakatauwehea atu x i te 3x^{2}-4x.

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

Whakatauwehea atu te kīanga pātahi 3x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

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

Me tuhi anō te kīanga whakatauwehe katoa.

$$\left(3x-4\right)\left(x+1\right)\left(2x+5\right)$$

Aromātai

$$\left(3x-4\right)\left(x+1\right)\left(2x+5\right)$$