3b^2-2b-5 eching

$$3 b ^ { 2 } - 2 b - 5$$

Omil

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

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3b^{2}+pb+qb-5 sifatida qayta yozilishi kerak. p va q ni topish uchun yechiladigan tizimni sozlang.

$$p+q=-2 pq=3\left(-5\right)=-15$$

pq manfiy boʻlganda, p va q da qarama-qarshi belgilar bor. p+q manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -15-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-15 3,-5$$

Har bir juftlik yigʻindisini hisoblang.

$$1-15=-14 3-5=-2$$

Yechim – -2 yigʻindisini beruvchi juftlik.

$$p=-5 q=3$$

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

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

3b^{2}-5b ichida b ni ajrating.

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

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

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

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.

$$3b^{2}-2b-5=0$$

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.

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

-2 kvadratini chiqarish.

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

-4 ni 3 marotabaga ko'paytirish.

$$b=\frac{-\left(-2\right)±\sqrt{4-12\left(-5\right)}}{2\times 3}$$

-12 ni -5 marotabaga ko'paytirish.

$$b=\frac{-\left(-2\right)±\sqrt{4+60}}{2\times 3}$$

4 ni 60 ga qo'shish.

$$b=\frac{-\left(-2\right)±\sqrt{64}}{2\times 3}$$

64 ning kvadrat ildizini chiqarish.

$$b=\frac{-\left(-2\right)±8}{2\times 3}$$

-2 ning teskarisi 2 ga teng.

$$b=\frac{2±8}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$b=\frac{2±8}{6}$$

b=\frac{2±8}{6} tenglamasini yeching, bunda ± musbat. 2 ni 8 ga qo'shish.

$$b=\frac{10}{6}$$

\frac{10}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

b=\frac{2±8}{6} tenglamasini yeching, bunda ± manfiy. 2 dan 8 ni ayirish.

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

-6 ni 6 ga bo'lish.

$$b=-1$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{5}{3} ga va x_{2} uchun -1 ga bo‘ling.

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

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

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

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

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

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

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

Baholash

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