12-5x-2x^2 eching

$$12 - 5 x - 2 x ^ { 2 }$$

Omil

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

Yechim qadamlari

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$-2x^{2}-5x+12$$

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -2x^{2}+ax+bx+12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-5 ab=-2\times 12=-24$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga 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 – -5 yigʻindisini beruvchi juftlik.

$$a=3 b=-8$$

-2x^{2}-5x+12 ni \left(-2x^{2}+3x\right)+\left(-8x+12\right) sifatida qaytadan yozish.

$$\left(-2x^{2}+3x\right)+\left(-8x+12\right)$$

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

$$-x\left(2x-3\right)-4\left(2x-3\right)$$

Distributiv funktsiyasidan foydalangan holda 2x-3 umumiy terminini chiqaring.

$$\left(2x-3\right)\left(-x-4\right)$$

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.

$$-2x^{2}-5x+12=0$$

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.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\times 12}}{2\left(-2\right)}$$

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 12}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\times 12}}{2\left(-2\right)}$$

8 ni 12 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+96}}{2\left(-2\right)}$$

25 ni 96 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{121}}{2\left(-2\right)}$$

121 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±11}{2\left(-2\right)}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±11}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{5±11}{-4}$$

x=\frac{5±11}{-4} tenglamasini yeching, bunda ± musbat. 5 ni 11 ga qo'shish.

$$x=\frac{16}{-4}$$

16 ni -4 ga bo'lish.

$$x=-4$$

x=\frac{5±11}{-4} tenglamasini yeching, bunda ± manfiy. 5 dan 11 ni ayirish.

$$x=-\frac{6}{-4}$$

\frac{-6}{-4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{3}{2}$$

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 \frac{3}{2} ga bo‘ling.

$$-2x^{2}-5x+12=-2\left(x-\left(-4\right)\right)\left(x-\frac{3}{2}\right)$$

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

$$-2x^{2}-5x+12=-2\left(x+4\right)\left(x-\frac{3}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{2} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$-2x^{2}-5x+12=-2\left(x+4\right)\times \frac{-2x+3}{-2}$$

-2 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

$$-2x^{2}-5x+12=\left(x+4\right)\left(-2x+3\right)$$

Baholash

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