-4x^2-12x+16 eching

$$- 4 x ^ { 2 } - 12 x + 16$$

Omil

$$4\left(1-x\right)\left(x+4\right)$$

Yechim qadamlari

4 omili.

$$4\left(-x^{2}-3x+4\right)$$

Hisoblang: -x^{2}-3x+4. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -x^{2}+ax+bx+4 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=-4=-4$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -4-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-4 2,-2$$

Har bir juftlik yigʻindisini hisoblang.

$$1-4=-3 2-2=0$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=1 b=-4$$

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

$$\left(-x^{2}+x\right)+\left(-4x+4\right)$$

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

$$x\left(-x+1\right)+4\left(-x+1\right)$$

Distributiv funktsiyasidan foydalangan holda -x+1 umumiy terminini chiqaring.

$$\left(-x+1\right)\left(x+4\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$4\left(-x+1\right)\left(x+4\right)$$

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.

$$-4x^{2}-12x+16=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-4\right)\times 16}}{2\left(-4\right)}$$

-12 kvadratini chiqarish.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-4\right)\times 16}}{2\left(-4\right)}$$

-4 ni -4 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+16\times 16}}{2\left(-4\right)}$$

16 ni 16 marotabaga ko'paytirish.

$$x=\frac{-\left(-12\right)±\sqrt{144+256}}{2\left(-4\right)}$$

144 ni 256 ga qo'shish.

$$x=\frac{-\left(-12\right)±\sqrt{400}}{2\left(-4\right)}$$

400 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-12\right)±20}{2\left(-4\right)}$$

-12 ning teskarisi 12 ga teng.

$$x=\frac{12±20}{2\left(-4\right)}$$

2 ni -4 marotabaga ko'paytirish.

$$x=\frac{12±20}{-8}$$

x=\frac{12±20}{-8} tenglamasini yeching, bunda ± musbat. 12 ni 20 ga qo'shish.

$$x=\frac{32}{-8}$$

32 ni -8 ga bo'lish.

$$x=-4$$

x=\frac{12±20}{-8} tenglamasini yeching, bunda ± manfiy. 12 dan 20 ni ayirish.

$$x=-\frac{8}{-8}$$

-8 ni -8 ga bo'lish.

$$x=1$$

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 1 ga bo‘ling.

$$-4x^{2}-12x+16=-4\left(x-\left(-4\right)\right)\left(x-1\right)$$

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

$$-4x^{2}-12x+16=-4\left(x+4\right)\left(x-1\right)$$

Baholash

$$4\left(1-x\right)\left(x+4\right)$$