6x^2-3x-3 eching

$$6 x ^ { 2 } - 3 x - 3$$

Omil

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

Yechim qadamlari

3 omili.

$$3\left(2x^{2}-x-1\right)$$

Hisoblang: 2x^{2}-x-1. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2x^{2}+ax+bx-1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=2\left(-1\right)=-2$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-2 b=1$$

2x^{2}-x-1 ni \left(2x^{2}-2x\right)+\left(x-1\right) sifatida qaytadan yozish.

$$\left(2x^{2}-2x\right)+\left(x-1\right)$$

2x^{2}-2x ichida 2x ni ajrating.

$$2x\left(x-1\right)+x-1$$

Distributiv funktsiyasidan foydalangan holda x-1 umumiy terminini chiqaring.

$$\left(x-1\right)\left(2x+1\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

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

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$6x^{2}-3x-3=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 6\left(-3\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9-24\left(-3\right)}}{2\times 6}$$

-24 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 6}$$

9 ni 72 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\times 6}$$

81 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±9}{2\times 6}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±9}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{3±9}{12}$$

x=\frac{3±9}{12} tenglamasini yeching, bunda ± musbat. 3 ni 9 ga qo'shish.

$$x=\frac{12}{12}$$

12 ni 12 ga bo'lish.

$$x=1$$

x=\frac{3±9}{12} tenglamasini yeching, bunda ± manfiy. 3 dan 9 ni ayirish.

$$x=-\frac{6}{12}$$

\frac{-6}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{1}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 1 ga va x_{2} uchun -\frac{1}{2} ga bo‘ling.

$$6x^{2}-3x-3=6\left(x-1\right)\left(x-\left(-\frac{1}{2}\right)\right)$$

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

$$6x^{2}-3x-3=6\left(x-1\right)\left(x+\frac{1}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6x^{2}-3x-3=6\left(x-1\right)\times \frac{2x+1}{2}$$

6 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

$$6x^{2}-3x-3=3\left(x-1\right)\left(2x+1\right)$$

Baholash

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