15x^2-26x-57 eching

$$15 { x }^{ 2 } -26x-57$$

Omil

$$\left(x-3\right)\left(15x+19\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 15x^{2}+ax+bx-57 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-26 ab=15\left(-57\right)=-855$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -855-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-855 3,-285 5,-171 9,-95 15,-57 19,-45$$

Har bir juftlik yigʻindisini hisoblang.

$$1-855=-854 3-285=-282 5-171=-166 9-95=-86 15-57=-42 19-45=-26$$

Yechim – -26 yigʻindisini beruvchi juftlik.

$$a=-45 b=19$$

15x^{2}-26x-57 ni \left(15x^{2}-45x\right)+\left(19x-57\right) sifatida qaytadan yozish.

$$\left(15x^{2}-45x\right)+\left(19x-57\right)$$

Birinchi guruhda 15x ni va ikkinchi guruhda 19 ni faktordan chiqaring.

$$15x\left(x-3\right)+19\left(x-3\right)$$

Distributiv funktsiyasidan foydalangan holda x-3 umumiy terminini chiqaring.

$$\left(x-3\right)\left(15x+19\right)$$

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.

$$15x^{2}-26x-57=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 15\left(-57\right)}}{2\times 15}$$

-26 kvadratini chiqarish.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 15\left(-57\right)}}{2\times 15}$$

-4 ni 15 marotabaga ko'paytirish.

$$x=\frac{-\left(-26\right)±\sqrt{676-60\left(-57\right)}}{2\times 15}$$

-60 ni -57 marotabaga ko'paytirish.

$$x=\frac{-\left(-26\right)±\sqrt{676+3420}}{2\times 15}$$

676 ni 3420 ga qo'shish.

$$x=\frac{-\left(-26\right)±\sqrt{4096}}{2\times 15}$$

4096 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-26\right)±64}{2\times 15}$$

-26 ning teskarisi 26 ga teng.

$$x=\frac{26±64}{2\times 15}$$

2 ni 15 marotabaga ko'paytirish.

$$x=\frac{26±64}{30}$$

x=\frac{26±64}{30} tenglamasini yeching, bunda ± musbat. 26 ni 64 ga qo'shish.

$$x=\frac{90}{30}$$

90 ni 30 ga bo'lish.

$$x=3$$

x=\frac{26±64}{30} tenglamasini yeching, bunda ± manfiy. 26 dan 64 ni ayirish.

$$x=-\frac{38}{30}$$

\frac{-38}{30} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{19}{15}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 3 ga va x_{2} uchun -\frac{19}{15} ga bo‘ling.

$$15x^{2}-26x-57=15\left(x-3\right)\left(x-\left(-\frac{19}{15}\right)\right)$$

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

$$15x^{2}-26x-57=15\left(x-3\right)\left(x+\frac{19}{15}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{19}{15} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$15x^{2}-26x-57=15\left(x-3\right)\times \frac{15x+19}{15}$$

15 va 15 ichida eng katta umumiy 15 faktorini bekor qiling.

$$15x^{2}-26x-57=\left(x-3\right)\left(15x+19\right)$$

Baholash

$$\left(x-3\right)\left(15x+19\right)$$