5x^2+23x+12 eching

$$5 { x }^{ 2 } +23x+12$$

Omil

$$\left(x+4\right)\left(5x+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 5x^{2}+ax+bx+12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=23 ab=5\times 12=60$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 60-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Har bir juftlik yigʻindisini hisoblang.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

Yechim – 23 yigʻindisini beruvchi juftlik.

$$a=3 b=20$$

5x^{2}+23x+12 ni \left(5x^{2}+3x\right)+\left(20x+12\right) sifatida qaytadan yozish.

$$\left(5x^{2}+3x\right)+\left(20x+12\right)$$

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

$$x\left(5x+3\right)+4\left(5x+3\right)$$

Distributiv funktsiyasidan foydalangan holda 5x+3 umumiy terminini chiqaring.

$$\left(5x+3\right)\left(x+4\right)$$

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.

$$5x^{2}+23x+12=0$$

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.

$$x=\frac{-23±\sqrt{23^{2}-4\times 5\times 12}}{2\times 5}$$

23 kvadratini chiqarish.

$$x=\frac{-23±\sqrt{529-4\times 5\times 12}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-23±\sqrt{529-20\times 12}}{2\times 5}$$

-20 ni 12 marotabaga ko'paytirish.

$$x=\frac{-23±\sqrt{529-240}}{2\times 5}$$

529 ni -240 ga qo'shish.

$$x=\frac{-23±\sqrt{289}}{2\times 5}$$

289 ning kvadrat ildizini chiqarish.

$$x=\frac{-23±17}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-23±17}{10}$$

x=\frac{-23±17}{10} tenglamasini yeching, bunda ± musbat. -23 ni 17 ga qo'shish.

$$x=-\frac{6}{10}$$

\frac{-6}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{3}{5}$$

x=\frac{-23±17}{10} tenglamasini yeching, bunda ± manfiy. -23 dan 17 ni ayirish.

$$x=-\frac{40}{10}$$

-40 ni 10 ga bo'lish.

$$x=-4$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{3}{5} ga va x_{2} uchun -4 ga bo‘ling.

$$5x^{2}+23x+12=5\left(x-\left(-\frac{3}{5}\right)\right)\left(x-\left(-4\right)\right)$$

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

$$5x^{2}+23x+12=5\left(x+\frac{3}{5}\right)\left(x+4\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{5} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$5x^{2}+23x+12=5\times \frac{5x+3}{5}\left(x+4\right)$$

5 va 5 ichida eng katta umumiy 5 faktorini bekor qiling.

$$5x^{2}+23x+12=\left(5x+3\right)\left(x+4\right)$$

Baholash

$$\left(x+4\right)\left(5x+3\right)$$