6x^2-x-40 eching

$$6 x ^ { 2 } - x - 40$$

Omil

$$\left(3x-8\right)\left(2x+5\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 6x^{2}+ax+bx-40 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=6\left(-40\right)=-240$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -240-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-240 2,-120 3,-80 4,-60 5,-48 6,-40 8,-30 10,-24 12,-20 15,-16$$

Har bir juftlik yigʻindisini hisoblang.

$$1-240=-239 2-120=-118 3-80=-77 4-60=-56 5-48=-43 6-40=-34 8-30=-22 10-24=-14 12-20=-8 15-16=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=-16 b=15$$

6x^{2}-x-40 ni \left(6x^{2}-16x\right)+\left(15x-40\right) sifatida qaytadan yozish.

$$\left(6x^{2}-16x\right)+\left(15x-40\right)$$

Birinchi guruhda 2x ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$2x\left(3x-8\right)+5\left(3x-8\right)$$

Distributiv funktsiyasidan foydalangan holda 3x-8 umumiy terminini chiqaring.

$$\left(3x-8\right)\left(2x+5\right)$$

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.

$$6x^{2}-x-40=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-40\right)}}{2\times 6}$$

-4 ni 6 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\left(-40\right)}}{2\times 6}$$

-24 ni -40 marotabaga ko'paytirish.

$$x=\frac{-\left(-1\right)±\sqrt{1+960}}{2\times 6}$$

1 ni 960 ga qo'shish.

$$x=\frac{-\left(-1\right)±\sqrt{961}}{2\times 6}$$

961 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-1\right)±31}{2\times 6}$$

-1 ning teskarisi 1 ga teng.

$$x=\frac{1±31}{2\times 6}$$

2 ni 6 marotabaga ko'paytirish.

$$x=\frac{1±31}{12}$$

x=\frac{1±31}{12} tenglamasini yeching, bunda ± musbat. 1 ni 31 ga qo'shish.

$$x=\frac{32}{12}$$

\frac{32}{12} ulushini 4 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{8}{3}$$

x=\frac{1±31}{12} tenglamasini yeching, bunda ± manfiy. 1 dan 31 ni ayirish.

$$x=-\frac{30}{12}$$

\frac{-30}{12} ulushini 6 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{8}{3} ga va x_{2} uchun -\frac{5}{2} ga bo‘ling.

$$6x^{2}-x-40=6\left(x-\frac{8}{3}\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$6x^{2}-x-40=6\left(x-\frac{8}{3}\right)\left(x+\frac{5}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{8}{3} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$6x^{2}-x-40=6\times \frac{3x-8}{3}\left(x+\frac{5}{2}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{5}{2} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$6x^{2}-x-40=6\times \frac{3x-8}{3}\times \frac{2x+5}{2}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{3x-8}{3} ni \frac{2x+5}{2} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$6x^{2}-x-40=6\times \frac{\left(3x-8\right)\left(2x+5\right)}{3\times 2}$$

3 ni 2 marotabaga ko'paytirish.

$$6x^{2}-x-40=6\times \frac{\left(3x-8\right)\left(2x+5\right)}{6}$$

6 va 6 ichida eng katta umumiy 6 faktorini bekor qiling.

$$6x^{2}-x-40=\left(3x-8\right)\left(2x+5\right)$$

Baholash

$$\left(3x-8\right)\left(2x+5\right)$$