3y^2+2y-8 eching

$$3 y ^ { 2 } + 2 y - 8$$

Omil

$$\left(3y-4\right)\left(y+2\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3y^{2}+ay+by-8 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=2 ab=3\left(-8\right)=-24$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -24-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

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

Har bir juftlik yigʻindisini hisoblang.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

Yechim – 2 yigʻindisini beruvchi juftlik.

$$a=-4 b=6$$

3y^{2}+2y-8 ni \left(3y^{2}-4y\right)+\left(6y-8\right) sifatida qaytadan yozish.

$$\left(3y^{2}-4y\right)+\left(6y-8\right)$$

Birinchi guruhda y ni va ikkinchi guruhda 2 ni faktordan chiqaring.

$$y\left(3y-4\right)+2\left(3y-4\right)$$

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

$$\left(3y-4\right)\left(y+2\right)$$

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.

$$3y^{2}+2y-8=0$$

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.

$$y=\frac{-2±\sqrt{2^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

2 kvadratini chiqarish.

$$y=\frac{-2±\sqrt{4-4\times 3\left(-8\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$y=\frac{-2±\sqrt{4-12\left(-8\right)}}{2\times 3}$$

-12 ni -8 marotabaga ko'paytirish.

$$y=\frac{-2±\sqrt{4+96}}{2\times 3}$$

4 ni 96 ga qo'shish.

$$y=\frac{-2±\sqrt{100}}{2\times 3}$$

100 ning kvadrat ildizini chiqarish.

$$y=\frac{-2±10}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$y=\frac{-2±10}{6}$$

y=\frac{-2±10}{6} tenglamasini yeching, bunda ± musbat. -2 ni 10 ga qo'shish.

$$y=\frac{8}{6}$$

\frac{8}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

y=\frac{-2±10}{6} tenglamasini yeching, bunda ± manfiy. -2 dan 10 ni ayirish.

$$y=-\frac{12}{6}$$

-12 ni 6 ga bo'lish.

$$y=-2$$

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

$$3y^{2}+2y-8=3\left(y-\frac{4}{3}\right)\left(y-\left(-2\right)\right)$$

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

$$3y^{2}+2y-8=3\left(y-\frac{4}{3}\right)\left(y+2\right)$$

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

$$3y^{2}+2y-8=3\times \frac{3y-4}{3}\left(y+2\right)$$

3 va 3 ichida eng katta umumiy 3 faktorini bekor qiling.

$$3y^{2}+2y-8=\left(3y-4\right)\left(y+2\right)$$

Baholash

$$\left(3y-4\right)\left(y+2\right)$$