f(x)=-3x^2-14x+24 eching

$$f ( x ) = - 3 x ^ { 2 } - 14 x + 24$$

Omil

$$-\left(3x-4\right)\left(x+6\right)$$

Yechim qadamlari

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

$$a+b=-14 ab=-3\times 24=-72$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga 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 – -14 yigʻindisini beruvchi juftlik.

$$a=4 b=-18$$

-3x^{2}-14x+24 ni \left(-3x^{2}+4x\right)+\left(-18x+24\right) sifatida qaytadan yozish.

$$\left(-3x^{2}+4x\right)+\left(-18x+24\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -6 ni faktordan chiqaring.

$$-x\left(3x-4\right)-6\left(3x-4\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-4 umumiy terminini chiqaring.

$$\left(3x-4\right)\left(-x-6\right)$$

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.

$$-3x^{2}-14x+24=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\left(-3\right)\times 24}}{2\left(-3\right)}$$

-14 kvadratini chiqarish.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\left(-3\right)\times 24}}{2\left(-3\right)}$$

-4 ni -3 marotabaga ko'paytirish.

$$x=\frac{-\left(-14\right)±\sqrt{196+12\times 24}}{2\left(-3\right)}$$

12 ni 24 marotabaga ko'paytirish.

$$x=\frac{-\left(-14\right)±\sqrt{196+288}}{2\left(-3\right)}$$

196 ni 288 ga qo'shish.

$$x=\frac{-\left(-14\right)±\sqrt{484}}{2\left(-3\right)}$$

484 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-14\right)±22}{2\left(-3\right)}$$

-14 ning teskarisi 14 ga teng.

$$x=\frac{14±22}{2\left(-3\right)}$$

2 ni -3 marotabaga ko'paytirish.

$$x=\frac{14±22}{-6}$$

x=\frac{14±22}{-6} tenglamasini yeching, bunda ± musbat. 14 ni 22 ga qo'shish.

$$x=\frac{36}{-6}$$

36 ni -6 ga bo'lish.

$$x=-6$$

x=\frac{14±22}{-6} tenglamasini yeching, bunda ± manfiy. 14 dan 22 ni ayirish.

$$x=-\frac{8}{-6}$$

\frac{-8}{-6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{4}{3}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -6 ga va x_{2} uchun \frac{4}{3} ga bo‘ling.

$$-3x^{2}-14x+24=-3\left(x-\left(-6\right)\right)\left(x-\frac{4}{3}\right)$$

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

$$-3x^{2}-14x+24=-3\left(x+6\right)\left(x-\frac{4}{3}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{4}{3} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$-3x^{2}-14x+24=-3\left(x+6\right)\times \frac{-3x+4}{-3}$$

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

$$-3x^{2}-14x+24=\left(x+6\right)\left(-3x+4\right)$$

Baholash

$$-\left(3x-4\right)\left(x+6\right)$$