2n^2-3n-20 eching

$$2 n ^ { 2 } - 3 n - 20$$

Omil

$$\left(n-4\right)\left(2n+5\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2n^{2}+an+bn-20 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=2\left(-20\right)=-40$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -40-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-40 2,-20 4,-10 5,-8$$

Har bir juftlik yigʻindisini hisoblang.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=-8 b=5$$

2n^{2}-3n-20 ni \left(2n^{2}-8n\right)+\left(5n-20\right) sifatida qaytadan yozish.

$$\left(2n^{2}-8n\right)+\left(5n-20\right)$$

Birinchi guruhda 2n ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$2n\left(n-4\right)+5\left(n-4\right)$$

Distributiv funktsiyasidan foydalangan holda n-4 umumiy terminini chiqaring.

$$\left(n-4\right)\left(2n+5\right)$$

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.

$$2n^{2}-3n-20=0$$

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.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

-3 kvadratini chiqarish.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-20\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$n=\frac{-\left(-3\right)±\sqrt{9-8\left(-20\right)}}{2\times 2}$$

-8 ni -20 marotabaga ko'paytirish.

$$n=\frac{-\left(-3\right)±\sqrt{9+160}}{2\times 2}$$

9 ni 160 ga qo'shish.

$$n=\frac{-\left(-3\right)±\sqrt{169}}{2\times 2}$$

169 ning kvadrat ildizini chiqarish.

$$n=\frac{-\left(-3\right)±13}{2\times 2}$$

-3 ning teskarisi 3 ga teng.

$$n=\frac{3±13}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$n=\frac{3±13}{4}$$

n=\frac{3±13}{4} tenglamasini yeching, bunda ± musbat. 3 ni 13 ga qo'shish.

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

16 ni 4 ga bo'lish.

$$n=4$$

n=\frac{3±13}{4} tenglamasini yeching, bunda ± manfiy. 3 dan 13 ni ayirish.

$$n=-\frac{10}{4}$$

\frac{-10}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$n=-\frac{5}{2}$$

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

$$2n^{2}-3n-20=2\left(n-4\right)\left(n-\left(-\frac{5}{2}\right)\right)$$

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

$$2n^{2}-3n-20=2\left(n-4\right)\left(n+\frac{5}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni n ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$2n^{2}-3n-20=2\left(n-4\right)\times \frac{2n+5}{2}$$

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

$$2n^{2}-3n-20=\left(n-4\right)\left(2n+5\right)$$

Baholash

$$\left(n-4\right)\left(2n+5\right)$$