3y^2+y-24 eching

$$3 y ^ { 2 } + y - 24$$

Omil

$$\left(3y-8\right)\left(y+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3y^{2}+ay+by-24 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=1 ab=3\left(-24\right)=-72$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. -72-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Har bir juftlik yigʻindisini hisoblang.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

Yechim – 1 yigʻindisini beruvchi juftlik.

$$a=-8 b=9$$

3y^{2}+y-24 ni \left(3y^{2}-8y\right)+\left(9y-24\right) sifatida qaytadan yozish.

$$\left(3y^{2}-8y\right)+\left(9y-24\right)$$

Birinchi guruhda y ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$y\left(3y-8\right)+3\left(3y-8\right)$$

Distributiv funktsiyasidan foydalangan holda 3y-8 umumiy terminini chiqaring.

$$\left(3y-8\right)\left(y+3\right)$$

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.

$$3y^{2}+y-24=0$$

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.

$$y=\frac{-1±\sqrt{1^{2}-4\times 3\left(-24\right)}}{2\times 3}$$

1 kvadratini chiqarish.

$$y=\frac{-1±\sqrt{1-4\times 3\left(-24\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$y=\frac{-1±\sqrt{1-12\left(-24\right)}}{2\times 3}$$

-12 ni -24 marotabaga ko'paytirish.

$$y=\frac{-1±\sqrt{1+288}}{2\times 3}$$

1 ni 288 ga qo'shish.

$$y=\frac{-1±\sqrt{289}}{2\times 3}$$

289 ning kvadrat ildizini chiqarish.

$$y=\frac{-1±17}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$y=\frac{-1±17}{6}$$

y=\frac{-1±17}{6} tenglamasini yeching, bunda ± musbat. -1 ni 17 ga qo'shish.

$$y=\frac{16}{6}$$

\frac{16}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$y=\frac{8}{3}$$

y=\frac{-1±17}{6} tenglamasini yeching, bunda ± manfiy. -1 dan 17 ni ayirish.

$$y=-\frac{18}{6}$$

-18 ni 6 ga bo'lish.

$$y=-3$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{8}{3} ga va x_{2} uchun -3 ga bo‘ling.

$$3y^{2}+y-24=3\left(y-\frac{8}{3}\right)\left(y-\left(-3\right)\right)$$

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

$$3y^{2}+y-24=3\left(y-\frac{8}{3}\right)\left(y+3\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{8}{3} ni y dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$3y^{2}+y-24=3\times \frac{3y-8}{3}\left(y+3\right)$$

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

$$3y^{2}+y-24=\left(3y-8\right)\left(y+3\right)$$

Baholash

$$\left(3y-8\right)\left(y+3\right)$$