-6x^2-2x+20 eching

$$-6 { x }^{ 2 } -2x+20$$

Omil

$$2\left(-x-2\right)\left(3x-5\right)$$

Yechim qadamlari

2 omili.

$$2\left(-3x^{2}-x+10\right)$$

Hisoblang: -3x^{2}-x+10. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -3x^{2}+ax+bx+10 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=-3\times 10=-30$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -30-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-30 2,-15 3,-10 5,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=5 b=-6$$

-3x^{2}-x+10 ni \left(-3x^{2}+5x\right)+\left(-6x+10\right) sifatida qaytadan yozish.

$$\left(-3x^{2}+5x\right)+\left(-6x+10\right)$$

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

$$-x\left(3x-5\right)-2\left(3x-5\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-5 umumiy terminini chiqaring.

$$\left(3x-5\right)\left(-x-2\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$2\left(3x-5\right)\left(-x-2\right)$$

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.

$$-6x^{2}-2x+20=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-6\right)\times 20}}{2\left(-6\right)}$$

-2 kvadratini chiqarish.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-6\right)\times 20}}{2\left(-6\right)}$$

-4 ni -6 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+24\times 20}}{2\left(-6\right)}$$

24 ni 20 marotabaga ko'paytirish.

$$x=\frac{-\left(-2\right)±\sqrt{4+480}}{2\left(-6\right)}$$

4 ni 480 ga qo'shish.

$$x=\frac{-\left(-2\right)±\sqrt{484}}{2\left(-6\right)}$$

484 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-2\right)±22}{2\left(-6\right)}$$

-2 ning teskarisi 2 ga teng.

$$x=\frac{2±22}{2\left(-6\right)}$$

2 ni -6 marotabaga ko'paytirish.

$$x=\frac{2±22}{-12}$$

x=\frac{2±22}{-12} tenglamasini yeching, bunda ± musbat. 2 ni 22 ga qo'shish.

$$x=\frac{24}{-12}$$

24 ni -12 ga bo'lish.

$$x=-2$$

x=\frac{2±22}{-12} tenglamasini yeching, bunda ± manfiy. 2 dan 22 ni ayirish.

$$x=-\frac{20}{-12}$$

\frac{-20}{-12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{5}{3}$$

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{5}{3} ga bo‘ling.

$$-6x^{2}-2x+20=-6\left(x-\left(-2\right)\right)\left(x-\frac{5}{3}\right)$$

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

$$-6x^{2}-2x+20=-6\left(x+2\right)\left(x-\frac{5}{3}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{5}{3} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$-6x^{2}-2x+20=-6\left(x+2\right)\times \frac{-3x+5}{-3}$$

-6 va 3 ichida eng katta umumiy 3 faktorini bekor qiling.

$$-6x^{2}-2x+20=2\left(x+2\right)\left(-3x+5\right)$$

Baholash

$$20-2x-6x^{2}$$