-10x^2-11x+6 eching

$$- 10 x ^ { 2 } - 11 x + 6$$

Omil

$$\left(2-5x\right)\left(2x+3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -10x^{2}+ax+bx+6 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-11 ab=-10\times 6=-60$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -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.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

Yechim – -11 yigʻindisini beruvchi juftlik.

$$a=4 b=-15$$

-10x^{2}-11x+6 ni \left(-10x^{2}+4x\right)+\left(-15x+6\right) sifatida qaytadan yozish.

$$\left(-10x^{2}+4x\right)+\left(-15x+6\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$2x\left(-5x+2\right)+3\left(-5x+2\right)$$

Distributiv funktsiyasidan foydalangan holda -5x+2 umumiy terminini chiqaring.

$$\left(-5x+2\right)\left(2x+3\right)$$

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.

$$-10x^{2}-11x+6=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-10\right)\times 6}}{2\left(-10\right)}$$

-11 kvadratini chiqarish.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-10\right)\times 6}}{2\left(-10\right)}$$

-4 ni -10 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121+40\times 6}}{2\left(-10\right)}$$

40 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\left(-10\right)}$$

121 ni 240 ga qo'shish.

$$x=\frac{-\left(-11\right)±\sqrt{361}}{2\left(-10\right)}$$

361 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-11\right)±19}{2\left(-10\right)}$$

-11 ning teskarisi 11 ga teng.

$$x=\frac{11±19}{2\left(-10\right)}$$

2 ni -10 marotabaga ko'paytirish.

$$x=\frac{11±19}{-20}$$

x=\frac{11±19}{-20} tenglamasini yeching, bunda ± musbat. 11 ni 19 ga qo'shish.

$$x=\frac{30}{-20}$$

\frac{30}{-20} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{3}{2}$$

x=\frac{11±19}{-20} tenglamasini yeching, bunda ± manfiy. 11 dan 19 ni ayirish.

$$x=-\frac{8}{-20}$$

\frac{-8}{-20} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{2}{5}$$

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

$$-10x^{2}-11x+6=-10\left(x-\left(-\frac{3}{2}\right)\right)\left(x-\frac{2}{5}\right)$$

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

$$-10x^{2}-11x+6=-10\left(x+\frac{3}{2}\right)\left(x-\frac{2}{5}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{3}{2} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$-10x^{2}-11x+6=-10\times \frac{-2x-3}{-2}\left(x-\frac{2}{5}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{2}{5} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$-10x^{2}-11x+6=-10\times \frac{-2x-3}{-2}\times \frac{-5x+2}{-5}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{-2x-3}{-2} ni \frac{-5x+2}{-5} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$-10x^{2}-11x+6=-10\times \frac{\left(-2x-3\right)\left(-5x+2\right)}{-2\left(-5\right)}$$

-2 ni -5 marotabaga ko'paytirish.

$$-10x^{2}-11x+6=-10\times \frac{\left(-2x-3\right)\left(-5x+2\right)}{10}$$

-10 va 10 ichida eng katta umumiy 10 faktorini bekor qiling.

$$-10x^{2}-11x+6=-\left(-2x-3\right)\left(-5x+2\right)$$

Baholash

$$\left(2-5x\right)\left(2x+3\right)$$