-(x^2)+2(x)+(3) eching

$$- ( x ^ { 2 } ) + 2 ( x ) + ( 3 )$$

Omil

$$-\left(x-3\right)\left(x+1\right)$$

Yechim qadamlari

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

$$a+b=2 ab=-3=-3$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=3 b=-1$$

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

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

Birinchi guruhda -x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$-x\left(x-3\right)-\left(x-3\right)$$

Distributiv funktsiyasidan foydalangan holda x-3 umumiy terminini chiqaring.

$$\left(x-3\right)\left(-x-1\right)$$

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.

$$-x^{2}+2x+3=0$$

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.

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

2 kvadratini chiqarish.

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

-4 ni -1 marotabaga ko'paytirish.

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

4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-2±\sqrt{4+12}}{2\left(-1\right)}$$

4 ni 12 ga qo'shish.

$$x=\frac{-2±\sqrt{16}}{2\left(-1\right)}$$

16 ning kvadrat ildizini chiqarish.

$$x=\frac{-2±4}{2\left(-1\right)}$$

2 ni -1 marotabaga ko'paytirish.

$$x=\frac{-2±4}{-2}$$

x=\frac{-2±4}{-2} tenglamasini yeching, bunda ± musbat. -2 ni 4 ga qo'shish.

$$x=\frac{2}{-2}$$

2 ni -2 ga bo'lish.

$$x=-1$$

x=\frac{-2±4}{-2} tenglamasini yeching, bunda ± manfiy. -2 dan 4 ni ayirish.

$$x=-\frac{6}{-2}$$

-6 ni -2 ga bo'lish.

$$x=3$$

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

$$-x^{2}+2x+3=-\left(x-\left(-1\right)\right)\left(x-3\right)$$

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

$$-x^{2}+2x+3=-\left(x+1\right)\left(x-3\right)$$

Baholash

$$-\left(x-3\right)\left(x+1\right)$$