2n^2+11n+12 eching

$$2 n ^ { 2 } + 11 n + 12$$

Omil

$$\left(n+4\right)\left(2n+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2n^{2}+an+bn+12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=11 ab=2\times 12=24$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 24-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,24 2,12 3,8 4,6$$

Har bir juftlik yigʻindisini hisoblang.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

Yechim – 11 yigʻindisini beruvchi juftlik.

$$a=3 b=8$$

2n^{2}+11n+12 ni \left(2n^{2}+3n\right)+\left(8n+12\right) sifatida qaytadan yozish.

$$\left(2n^{2}+3n\right)+\left(8n+12\right)$$

Birinchi guruhda n ni va ikkinchi guruhda 4 ni faktordan chiqaring.

$$n\left(2n+3\right)+4\left(2n+3\right)$$

Distributiv funktsiyasidan foydalangan holda 2n+3 umumiy terminini chiqaring.

$$\left(2n+3\right)\left(n+4\right)$$

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.

$$2n^{2}+11n+12=0$$

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.

$$n=\frac{-11±\sqrt{11^{2}-4\times 2\times 12}}{2\times 2}$$

11 kvadratini chiqarish.

$$n=\frac{-11±\sqrt{121-4\times 2\times 12}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$n=\frac{-11±\sqrt{121-8\times 12}}{2\times 2}$$

-8 ni 12 marotabaga ko'paytirish.

$$n=\frac{-11±\sqrt{121-96}}{2\times 2}$$

121 ni -96 ga qo'shish.

$$n=\frac{-11±\sqrt{25}}{2\times 2}$$

25 ning kvadrat ildizini chiqarish.

$$n=\frac{-11±5}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$n=\frac{-11±5}{4}$$

n=\frac{-11±5}{4} tenglamasini yeching, bunda ± musbat. -11 ni 5 ga qo'shish.

$$n=-\frac{6}{4}$$

\frac{-6}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

n=\frac{-11±5}{4} tenglamasini yeching, bunda ± manfiy. -11 dan 5 ni ayirish.

$$n=-\frac{16}{4}$$

-16 ni 4 ga bo'lish.

$$n=-4$$

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

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

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

$$2n^{2}+11n+12=2\left(n+\frac{3}{2}\right)\left(n+4\right)$$

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

$$2n^{2}+11n+12=2\times \frac{2n+3}{2}\left(n+4\right)$$

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

$$2n^{2}+11n+12=\left(2n+3\right)\left(n+4\right)$$

Baholash

$$\left(n+4\right)\left(2n+3\right)$$