3x^2-17x-6 eching

$$3 x ^ { 2 } - 17 x - 6$$

Omil

$$\left(x-6\right)\left(3x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3x^{2}+ax+bx-6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-17 ab=3\left(-6\right)=-18$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -18-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-18 2,-9 3,-6$$

Har bir juftlik yigʻindisini hisoblang.

$$1-18=-17 2-9=-7 3-6=-3$$

Yechim – -17 yigʻindisini beruvchi juftlik.

$$a=-18 b=1$$

3x^{2}-17x-6 ni \left(3x^{2}-18x\right)+\left(x-6\right) sifatida qaytadan yozish.

$$\left(3x^{2}-18x\right)+\left(x-6\right)$$

3x^{2}-18x ichida 3x ni ajrating.

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

Distributiv funktsiyasidan foydalangan holda x-6 umumiy terminini chiqaring.

$$\left(x-6\right)\left(3x+1\right)$$

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.

$$3x^{2}-17x-6=0$$

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.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 3\left(-6\right)}}{2\times 3}$$

-17 kvadratini chiqarish.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 3\left(-6\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289-12\left(-6\right)}}{2\times 3}$$

-12 ni -6 marotabaga ko'paytirish.

$$x=\frac{-\left(-17\right)±\sqrt{289+72}}{2\times 3}$$

289 ni 72 ga qo'shish.

$$x=\frac{-\left(-17\right)±\sqrt{361}}{2\times 3}$$

361 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-17\right)±19}{2\times 3}$$

-17 ning teskarisi 17 ga teng.

$$x=\frac{17±19}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$x=\frac{17±19}{6}$$

x=\frac{17±19}{6} tenglamasini yeching, bunda ± musbat. 17 ni 19 ga qo'shish.

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

36 ni 6 ga bo'lish.

$$x=6$$

x=\frac{17±19}{6} tenglamasini yeching, bunda ± manfiy. 17 dan 19 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 6 ga va x_{2} uchun -\frac{1}{3} ga bo‘ling.

$$3x^{2}-17x-6=3\left(x-6\right)\left(x-\left(-\frac{1}{3}\right)\right)$$

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

$$3x^{2}-17x-6=3\left(x-6\right)\left(x+\frac{1}{3}\right)$$

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.

$$3x^{2}-17x-6=3\left(x-6\right)\times \frac{3x+1}{3}$$

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

$$3x^{2}-17x-6=\left(x-6\right)\left(3x+1\right)$$

Baholash

$$\left(x-6\right)\left(3x+1\right)$$