3x^2-53x+232= eching

$$3 { x }^{ 2 } -53x+232=$$

Omil

$$\left(x-8\right)\left(3x-29\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3x^{2}+ax+bx+232 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-53 ab=3\times 232=696$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 696-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-696 -2,-348 -3,-232 -4,-174 -6,-116 -8,-87 -12,-58 -24,-29$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-696=-697 -2-348=-350 -3-232=-235 -4-174=-178 -6-116=-122 -8-87=-95 -12-58=-70 -24-29=-53$$

Yechim – -53 yigʻindisini beruvchi juftlik.

$$a=-29 b=-24$$

3x^{2}-53x+232 ni \left(3x^{2}-29x\right)+\left(-24x+232\right) sifatida qaytadan yozish.

$$\left(3x^{2}-29x\right)+\left(-24x+232\right)$$

Birinchi guruhda x ni va ikkinchi guruhda -8 ni faktordan chiqaring.

$$x\left(3x-29\right)-8\left(3x-29\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-29 umumiy terminini chiqaring.

$$\left(3x-29\right)\left(x-8\right)$$

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.

$$3x^{2}-53x+232=0$$

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.

$$x=\frac{-\left(-53\right)±\sqrt{\left(-53\right)^{2}-4\times 3\times 232}}{2\times 3}$$

-53 kvadratini chiqarish.

$$x=\frac{-\left(-53\right)±\sqrt{2809-4\times 3\times 232}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-53\right)±\sqrt{2809-12\times 232}}{2\times 3}$$

-12 ni 232 marotabaga ko'paytirish.

$$x=\frac{-\left(-53\right)±\sqrt{2809-2784}}{2\times 3}$$

2809 ni -2784 ga qo'shish.

$$x=\frac{-\left(-53\right)±\sqrt{25}}{2\times 3}$$

25 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-53\right)±5}{2\times 3}$$

-53 ning teskarisi 53 ga teng.

$$x=\frac{53±5}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{53±5}{6}$$

x=\frac{53±5}{6} tenglamasini yeching, bunda ± musbat. 53 ni 5 ga qo'shish.

$$x=\frac{58}{6}$$

\frac{58}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{29}{3}$$

x=\frac{53±5}{6} tenglamasini yeching, bunda ± manfiy. 53 dan 5 ni ayirish.

$$x=\frac{48}{6}$$

48 ni 6 ga bo'lish.

$$x=8$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{29}{3} ga va x_{2} uchun 8 ga bo‘ling.

$$3x^{2}-53x+232=3\left(x-\frac{29}{3}\right)\left(x-8\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{29}{3} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$3x^{2}-53x+232=3\times \frac{3x-29}{3}\left(x-8\right)$$

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

$$3x^{2}-53x+232=\left(3x-29\right)\left(x-8\right)$$

Baholash

$$\left(x-8\right)\left(3x-29\right)$$