6a^2-11a-10 eching

$$6 a ^ { 2 } - 11 a - 10$$

Omil

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

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6a^{2}+pa+qa-10 sifatida qayta yozilishi kerak. p va q ni topish uchun yechiladigan tizimni sozlang.

$$p+q=-11 pq=6\left(-10\right)=-60$$

pq manfiy boʻlganda, p va q da qarama-qarshi belgilar bor. p+q manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -60-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$p=-15 q=4$$

6a^{2}-11a-10 ni \left(6a^{2}-15a\right)+\left(4a-10\right) sifatida qaytadan yozish.

$$\left(6a^{2}-15a\right)+\left(4a-10\right)$$

Birinchi guruhda 3a ni va ikkinchi guruhda 2 ni faktordan chiqaring.

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

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

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

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.

$$6a^{2}-11a-10=0$$

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.

$$a=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\left(-10\right)}}{2\times 6}$$

-11 kvadratini chiqarish.

$$a=\frac{-\left(-11\right)±\sqrt{121-4\times 6\left(-10\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$a=\frac{-\left(-11\right)±\sqrt{121-24\left(-10\right)}}{2\times 6}$$

-24 ni -10 marotabaga ko'paytirish.

$$a=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 6}$$

121 ni 240 ga qo'shish.

$$a=\frac{-\left(-11\right)±\sqrt{361}}{2\times 6}$$

361 ning kvadrat ildizini chiqarish.

$$a=\frac{-\left(-11\right)±19}{2\times 6}$$

-11 ning teskarisi 11 ga teng.

$$a=\frac{11±19}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$a=\frac{11±19}{12}$$

a=\frac{11±19}{12} tenglamasini yeching, bunda ± musbat. 11 ni 19 ga qo'shish.

$$a=\frac{30}{12}$$

\frac{30}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

a=\frac{11±19}{12} tenglamasini yeching, bunda ± manfiy. 11 dan 19 ni ayirish.

$$a=-\frac{8}{12}$$

\frac{-8}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$a=-\frac{2}{3}$$

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{2}{3} ga bo‘ling.

$$6a^{2}-11a-10=6\left(a-\frac{5}{2}\right)\left(a-\left(-\frac{2}{3}\right)\right)$$

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

$$6a^{2}-11a-10=6\left(a-\frac{5}{2}\right)\left(a+\frac{2}{3}\right)$$

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

$$6a^{2}-11a-10=6\times \frac{2a-5}{2}\left(a+\frac{2}{3}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{2}{3} ni a ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6a^{2}-11a-10=6\times \frac{2a-5}{2}\times \frac{3a+2}{3}$$

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

$$6a^{2}-11a-10=6\times \frac{\left(2a-5\right)\left(3a+2\right)}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$6a^{2}-11a-10=6\times \frac{\left(2a-5\right)\left(3a+2\right)}{6}$$

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

$$6a^{2}-11a-10=\left(2a-5\right)\left(3a+2\right)$$

Baholash

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