3x^2-20x-7 eching

$$3 x ^ { 2 } - 20 x - 7$$

Omil

$$\left(x-7\right)\left(3x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3x^{2}+ax+bx-7 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-20 ab=3\left(-7\right)=-21$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -21-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-21 3,-7$$

Har bir juftlik yigʻindisini hisoblang.

$$1-21=-20 3-7=-4$$

Yechim – -20 yigʻindisini beruvchi juftlik.

$$a=-21 b=1$$

3x^{2}-20x-7 ni \left(3x^{2}-21x\right)+\left(x-7\right) sifatida qaytadan yozish.

$$\left(3x^{2}-21x\right)+\left(x-7\right)$$

3x^{2}-21x ichida 3x ni ajrating.

$$3x\left(x-7\right)+x-7$$

Distributiv funktsiyasidan foydalangan holda x-7 umumiy terminini chiqaring.

$$\left(x-7\right)\left(3x+1\right)$$

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.

$$3x^{2}-20x-7=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\left(-7\right)}}{2\times 3}$$

-20 kvadratini chiqarish.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 3\left(-7\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400-12\left(-7\right)}}{2\times 3}$$

-12 ni -7 marotabaga ko'paytirish.

$$x=\frac{-\left(-20\right)±\sqrt{400+84}}{2\times 3}$$

400 ni 84 ga qo'shish.

$$x=\frac{-\left(-20\right)±\sqrt{484}}{2\times 3}$$

484 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-20\right)±22}{2\times 3}$$

-20 ning teskarisi 20 ga teng.

$$x=\frac{20±22}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{20±22}{6}$$

x=\frac{20±22}{6} tenglamasini yeching, bunda ± musbat. 20 ni 22 ga qo'shish.

$$x=\frac{42}{6}$$

42 ni 6 ga bo'lish.

$$x=7$$

x=\frac{20±22}{6} tenglamasini yeching, bunda ± manfiy. 20 dan 22 ni ayirish.

$$x=-\frac{2}{6}$$

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

$$x=-\frac{1}{3}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 7 ga va x_{2} uchun -\frac{1}{3} ga bo‘ling.

$$3x^{2}-20x-7=3\left(x-7\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$3x^{2}-20x-7=3\left(x-7\right)\left(x+\frac{1}{3}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{3} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$3x^{2}-20x-7=3\left(x-7\right)\times \frac{3x+1}{3}$$

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

$$3x^{2}-20x-7=\left(x-7\right)\left(3x+1\right)$$

Baholash

$$\left(x-7\right)\left(3x+1\right)$$