3b^2-22b-80 eching

$$3 b ^ { 2 } - 22 b - 80$$

Omil

$$\left(b-10\right)\left(3b+8\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3b^{2}+pb+qb-80 sifatida qayta yozilishi kerak. p va q ni topish uchun yechiladigan tizimni sozlang.

$$p+q=-22 pq=3\left(-80\right)=-240$$

pq manfiy boʻlganda, p va q da qarama-qarshi belgilar bor. p+q 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 – -22 yigʻindisini beruvchi juftlik.

$$p=-30 q=8$$

3b^{2}-22b-80 ni \left(3b^{2}-30b\right)+\left(8b-80\right) sifatida qaytadan yozish.

$$\left(3b^{2}-30b\right)+\left(8b-80\right)$$

Birinchi guruhda 3b ni va ikkinchi guruhda 8 ni faktordan chiqaring.

$$3b\left(b-10\right)+8\left(b-10\right)$$

Distributiv funktsiyasidan foydalangan holda b-10 umumiy terminini chiqaring.

$$\left(b-10\right)\left(3b+8\right)$$

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.

$$3b^{2}-22b-80=0$$

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.

$$b=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 3\left(-80\right)}}{2\times 3}$$

-22 kvadratini chiqarish.

$$b=\frac{-\left(-22\right)±\sqrt{484-4\times 3\left(-80\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$b=\frac{-\left(-22\right)±\sqrt{484-12\left(-80\right)}}{2\times 3}$$

-12 ni -80 marotabaga ko'paytirish.

$$b=\frac{-\left(-22\right)±\sqrt{484+960}}{2\times 3}$$

484 ni 960 ga qo'shish.

$$b=\frac{-\left(-22\right)±\sqrt{1444}}{2\times 3}$$

1444 ning kvadrat ildizini chiqarish.

$$b=\frac{-\left(-22\right)±38}{2\times 3}$$

-22 ning teskarisi 22 ga teng.

$$b=\frac{22±38}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$b=\frac{22±38}{6}$$

b=\frac{22±38}{6} tenglamasini yeching, bunda ± musbat. 22 ni 38 ga qo'shish.

$$b=\frac{60}{6}$$

60 ni 6 ga bo'lish.

$$b=10$$

b=\frac{22±38}{6} tenglamasini yeching, bunda ± manfiy. 22 dan 38 ni ayirish.

$$b=-\frac{16}{6}$$

\frac{-16}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$b=-\frac{8}{3}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 10 ga va x_{2} uchun -\frac{8}{3} ga bo‘ling.

$$3b^{2}-22b-80=3\left(b-10\right)\left(b-\left(-\frac{8}{3}\right)\right)$$

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

$$3b^{2}-22b-80=3\left(b-10\right)\left(b+\frac{8}{3}\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{8}{3} ni b ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$3b^{2}-22b-80=3\left(b-10\right)\times \frac{3b+8}{3}$$

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

$$3b^{2}-22b-80=\left(b-10\right)\left(3b+8\right)$$

Baholash

$$\left(b-10\right)\left(3b+8\right)$$