10c^2-19c-15 eching

$$10 c ^ { 2 } - 19 c - 15$$

Omil

$$\left(2c-5\right)\left(5c+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 10c^{2}+ac+bc-15 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-19 ab=10\left(-15\right)=-150$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -150-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Har bir juftlik yigʻindisini hisoblang.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

Yechim – -19 yigʻindisini beruvchi juftlik.

$$a=-25 b=6$$

10c^{2}-19c-15 ni \left(10c^{2}-25c\right)+\left(6c-15\right) sifatida qaytadan yozish.

$$\left(10c^{2}-25c\right)+\left(6c-15\right)$$

Birinchi guruhda 5c ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$5c\left(2c-5\right)+3\left(2c-5\right)$$

Distributiv funktsiyasidan foydalangan holda 2c-5 umumiy terminini chiqaring.

$$\left(2c-5\right)\left(5c+3\right)$$

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.

$$10c^{2}-19c-15=0$$

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.

$$c=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 10\left(-15\right)}}{2\times 10}$$

-19 kvadratini chiqarish.

$$c=\frac{-\left(-19\right)±\sqrt{361-4\times 10\left(-15\right)}}{2\times 10}$$

-4 ni 10 marotabaga ko'paytirish.

$$c=\frac{-\left(-19\right)±\sqrt{361-40\left(-15\right)}}{2\times 10}$$

-40 ni -15 marotabaga ko'paytirish.

$$c=\frac{-\left(-19\right)±\sqrt{361+600}}{2\times 10}$$

361 ni 600 ga qo'shish.

$$c=\frac{-\left(-19\right)±\sqrt{961}}{2\times 10}$$

961 ning kvadrat ildizini chiqarish.

$$c=\frac{-\left(-19\right)±31}{2\times 10}$$

-19 ning teskarisi 19 ga teng.

$$c=\frac{19±31}{2\times 10}$$

2 ni 10 marotabaga ko'paytirish.

$$c=\frac{19±31}{20}$$

c=\frac{19±31}{20} tenglamasini yeching, bunda ± musbat. 19 ni 31 ga qo'shish.

$$c=\frac{50}{20}$$

\frac{50}{20} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$c=\frac{5}{2}$$

c=\frac{19±31}{20} tenglamasini yeching, bunda ± manfiy. 19 dan 31 ni ayirish.

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

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

$$c=-\frac{3}{5}$$

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

$$10c^{2}-19c-15=10\left(c-\frac{5}{2}\right)\left(c-\left(-\frac{3}{5}\right)\right)$$

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

$$10c^{2}-19c-15=10\left(c-\frac{5}{2}\right)\left(c+\frac{3}{5}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{5}{2} ni c dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$10c^{2}-19c-15=10\times \frac{2c-5}{2}\left(c+\frac{3}{5}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{5} ni c ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$10c^{2}-19c-15=10\times \frac{2c-5}{2}\times \frac{5c+3}{5}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{2c-5}{2} ni \frac{5c+3}{5} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$10c^{2}-19c-15=10\times \frac{\left(2c-5\right)\left(5c+3\right)}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$10c^{2}-19c-15=10\times \frac{\left(2c-5\right)\left(5c+3\right)}{10}$$

10 va 10 ichida eng katta umumiy 10 faktorini bekor qiling.

$$10c^{2}-19c-15=\left(2c-5\right)\left(5c+3\right)$$

Baholash

$$\left(2c-5\right)\left(5c+3\right)$$