-3x^2-5x+2 eching

$$- 3 x ^ { 2 } - 5 x + 2$$

Omil

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

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -3x^{2}+ax+bx+2 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=-3\times 2=-6$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -6-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-6 2,-3$$

Har bir juftlik yigʻindisini hisoblang.

$$1-6=-5 2-3=-1$$

Yechim – -5 yigʻindisini beruvchi juftlik.

$$a=1 b=-6$$

-3x^{2}-5x+2 ni \left(-3x^{2}+x\right)+\left(-6x+2\right) sifatida qaytadan yozish.

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

Birinchi guruhda -x ni va ikkinchi guruhda -2 ni faktordan chiqaring.

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

Distributiv funktsiyasidan foydalangan holda 3x-1 umumiy terminini chiqaring.

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

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.

$$-3x^{2}-5x+2=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+12\times 2}}{2\left(-3\right)}$$

12 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\left(-3\right)}$$

25 ni 24 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\left(-3\right)}$$

49 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±7}{2\left(-3\right)}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±7}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{5±7}{-6}$$

x=\frac{5±7}{-6} tenglamasini yeching, bunda ± musbat. 5 ni 7 ga qo'shish.

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

12 ni -6 ga bo'lish.

$$x=-2$$

x=\frac{5±7}{-6} tenglamasini yeching, bunda ± manfiy. 5 dan 7 ni ayirish.

$$x=-\frac{2}{-6}$$

\frac{-2}{-6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{1}{3}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -2 ga va x_{2} uchun \frac{1}{3} ga bo‘ling.

$$-3x^{2}-5x+2=-3\left(x-\left(-2\right)\right)\left(x-\frac{1}{3}\right)$$

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

$$-3x^{2}-5x+2=-3\left(x+2\right)\left(x-\frac{1}{3}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{1}{3} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$-3x^{2}-5x+2=-3\left(x+2\right)\times \frac{-3x+1}{-3}$$

-3 va 3 ichida eng katta umumiy 3 faktorini bekor qiling.

$$-3x^{2}-5x+2=\left(x+2\right)\left(-3x+1\right)$$

Baholash

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