6b^2-27b-15 eching

$$6 b ^ { 2 } - 27 b - 15$$

Omil

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

Yechim qadamlari

3 omili.

$$3\left(2b^{2}-9b-5\right)$$

Hisoblang: 2b^{2}-9b-5. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2b^{2}+pb+qb-5 sifatida qayta yozilishi kerak. p va q ni topish uchun yechiladigan tizimni sozlang.

$$p+q=-9 pq=2\left(-5\right)=-10$$

pq manfiy boʻlganda, p va q da qarama-qarshi belgilar bor. p+q manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -10-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-10 2,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$1-10=-9 2-5=-3$$

Yechim – -9 yigʻindisini beruvchi juftlik.

$$p=-10 q=1$$

2b^{2}-9b-5 ni \left(2b^{2}-10b\right)+\left(b-5\right) sifatida qaytadan yozish.

$$\left(2b^{2}-10b\right)+\left(b-5\right)$$

2b^{2}-10b ichida 2b ni ajrating.

$$2b\left(b-5\right)+b-5$$

Distributiv funktsiyasidan foydalangan holda b-5 umumiy terminini chiqaring.

$$\left(b-5\right)\left(2b+1\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

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

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.

$$6b^{2}-27b-15=0$$

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.

$$b=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 6\left(-15\right)}}{2\times 6}$$

-27 kvadratini chiqarish.

$$b=\frac{-\left(-27\right)±\sqrt{729-4\times 6\left(-15\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$b=\frac{-\left(-27\right)±\sqrt{729-24\left(-15\right)}}{2\times 6}$$

-24 ni -15 marotabaga ko'paytirish.

$$b=\frac{-\left(-27\right)±\sqrt{729+360}}{2\times 6}$$

729 ni 360 ga qo'shish.

$$b=\frac{-\left(-27\right)±\sqrt{1089}}{2\times 6}$$

1089 ning kvadrat ildizini chiqarish.

$$b=\frac{-\left(-27\right)±33}{2\times 6}$$

-27 ning teskarisi 27 ga teng.

$$b=\frac{27±33}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$b=\frac{27±33}{12}$$

b=\frac{27±33}{12} tenglamasini yeching, bunda ± musbat. 27 ni 33 ga qo'shish.

$$b=\frac{60}{12}$$

60 ni 12 ga bo'lish.

$$b=5$$

b=\frac{27±33}{12} tenglamasini yeching, bunda ± manfiy. 27 dan 33 ni ayirish.

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

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

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 5 ga va x_{2} uchun -\frac{1}{2} ga bo‘ling.

$$6b^{2}-27b-15=6\left(b-5\right)\left(b-\left(-\frac{1}{2}\right)\right)$$

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

$$6b^{2}-27b-15=6\left(b-5\right)\left(b+\frac{1}{2}\right)$$

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

$$6b^{2}-27b-15=6\left(b-5\right)\times \frac{2b+1}{2}$$

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

$$6b^{2}-27b-15=3\left(b-5\right)\left(2b+1\right)$$

Baholash

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