12c^2+11c-15 eching

$$12 c ^ { 2 } + 11 c - 15$$

Omil

$$\left(4c-3\right)\left(3c+5\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 12c^{2}+ac+bc-15 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=11 ab=12\left(-15\right)=-180$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -180-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

Yechim – 11 yigʻindisini beruvchi juftlik.

$$a=-9 b=20$$

12c^{2}+11c-15 ni \left(12c^{2}-9c\right)+\left(20c-15\right) sifatida qaytadan yozish.

$$\left(12c^{2}-9c\right)+\left(20c-15\right)$$

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

$$3c\left(4c-3\right)+5\left(4c-3\right)$$

Distributiv funktsiyasidan foydalangan holda 4c-3 umumiy terminini chiqaring.

$$\left(4c-3\right)\left(3c+5\right)$$

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.

$$12c^{2}+11c-15=0$$

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.

$$c=\frac{-11±\sqrt{11^{2}-4\times 12\left(-15\right)}}{2\times 12}$$

11 kvadratini chiqarish.

$$c=\frac{-11±\sqrt{121-4\times 12\left(-15\right)}}{2\times 12}$$

-4 ni 12 marotabaga ko'paytirish.

$$c=\frac{-11±\sqrt{121-48\left(-15\right)}}{2\times 12}$$

-48 ni -15 marotabaga ko'paytirish.

$$c=\frac{-11±\sqrt{121+720}}{2\times 12}$$

121 ni 720 ga qo'shish.

$$c=\frac{-11±\sqrt{841}}{2\times 12}$$

841 ning kvadrat ildizini chiqarish.

$$c=\frac{-11±29}{2\times 12}$$

2 ni 12 marotabaga ko'paytirish.

$$c=\frac{-11±29}{24}$$

c=\frac{-11±29}{24} tenglamasini yeching, bunda ± musbat. -11 ni 29 ga qo'shish.

$$c=\frac{18}{24}$$

\frac{18}{24} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$c=\frac{3}{4}$$

c=\frac{-11±29}{24} tenglamasini yeching, bunda ± manfiy. -11 dan 29 ni ayirish.

$$c=-\frac{40}{24}$$

\frac{-40}{24} ulushini 8 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{3}{4} ga va x_{2} uchun -\frac{5}{3} ga bo‘ling.

$$12c^{2}+11c-15=12\left(c-\frac{3}{4}\right)\left(c-\left(-\frac{5}{3}\right)\right)$$

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

$$12c^{2}+11c-15=12\left(c-\frac{3}{4}\right)\left(c+\frac{5}{3}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{4} ni c dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$12c^{2}+11c-15=12\times \frac{4c-3}{4}\left(c+\frac{5}{3}\right)$$

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

$$12c^{2}+11c-15=12\times \frac{4c-3}{4}\times \frac{3c+5}{3}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{4c-3}{4} ni \frac{3c+5}{3} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$12c^{2}+11c-15=12\times \frac{\left(4c-3\right)\left(3c+5\right)}{4\times 3}$$

4 ni 3 marotabaga ko'paytirish.

$$12c^{2}+11c-15=12\times \frac{\left(4c-3\right)\left(3c+5\right)}{12}$$

12 va 12 ichida eng katta umumiy 12 faktorini bekor qiling.

$$12c^{2}+11c-15=\left(4c-3\right)\left(3c+5\right)$$

Baholash

$$\left(4c-3\right)\left(3c+5\right)$$