-5y^2-8y+4 eching

$$- 5 y ^ { 2 } - 8 y + 4$$

Omil

$$-\left(5y-2\right)\left(y+2\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -5y^{2}+ay+by+4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-8 ab=-5\times 4=-20$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -20-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

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

Har bir juftlik yigʻindisini hisoblang.

$$1-20=-19 2-10=-8 4-5=-1$$

Yechim – -8 yigʻindisini beruvchi juftlik.

$$a=2 b=-10$$

-5y^{2}-8y+4 ni \left(-5y^{2}+2y\right)+\left(-10y+4\right) sifatida qaytadan yozish.

$$\left(-5y^{2}+2y\right)+\left(-10y+4\right)$$

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

$$-y\left(5y-2\right)-2\left(5y-2\right)$$

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

$$\left(5y-2\right)\left(-y-2\right)$$

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.

$$-5y^{2}-8y+4=0$$

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.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-5\right)\times 4}}{2\left(-5\right)}$$

-8 kvadratini chiqarish.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\left(-5\right)\times 4}}{2\left(-5\right)}$$

-4 ni -5 marotabaga ko'paytirish.

$$y=\frac{-\left(-8\right)±\sqrt{64+20\times 4}}{2\left(-5\right)}$$

20 ni 4 marotabaga ko'paytirish.

$$y=\frac{-\left(-8\right)±\sqrt{64+80}}{2\left(-5\right)}$$

64 ni 80 ga qo'shish.

$$y=\frac{-\left(-8\right)±\sqrt{144}}{2\left(-5\right)}$$

144 ning kvadrat ildizini chiqarish.

$$y=\frac{-\left(-8\right)±12}{2\left(-5\right)}$$

-8 ning teskarisi 8 ga teng.

$$y=\frac{8±12}{2\left(-5\right)}$$

2 ni -5 marotabaga ko'paytirish.

$$y=\frac{8±12}{-10}$$

y=\frac{8±12}{-10} tenglamasini yeching, bunda ± musbat. 8 ni 12 ga qo'shish.

$$y=\frac{20}{-10}$$

20 ni -10 ga bo'lish.

$$y=-2$$

y=\frac{8±12}{-10} tenglamasini yeching, bunda ± manfiy. 8 dan 12 ni ayirish.

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

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

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

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

$$-5y^{2}-8y+4=-5\left(y-\left(-2\right)\right)\left(y-\frac{2}{5}\right)$$

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

$$-5y^{2}-8y+4=-5\left(y+2\right)\left(y-\frac{2}{5}\right)$$

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

$$-5y^{2}-8y+4=-5\left(y+2\right)\times \frac{-5y+2}{-5}$$

-5 va 5 ichida eng katta umumiy 5 faktorini bekor qiling.

$$-5y^{2}-8y+4=\left(y+2\right)\left(-5y+2\right)$$

Baholash

$$-\left(5y-2\right)\left(y+2\right)$$