3d^2+20d+12 eching

$$3 d ^ { 2 } + 20 d + 12$$

Omil

$$\left(d+6\right)\left(3d+2\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 3d^{2}+ad+bd+12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=20 ab=3\times 12=36$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b musbat boʻlganda, a va b ikkisi ham musbat. 36-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,36 2,18 3,12 4,9 6,6$$

Har bir juftlik yigʻindisini hisoblang.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

Yechim – 20 yigʻindisini beruvchi juftlik.

$$a=2 b=18$$

3d^{2}+20d+12 ni \left(3d^{2}+2d\right)+\left(18d+12\right) sifatida qaytadan yozish.

$$\left(3d^{2}+2d\right)+\left(18d+12\right)$$

Birinchi guruhda d ni va ikkinchi guruhda 6 ni faktordan chiqaring.

$$d\left(3d+2\right)+6\left(3d+2\right)$$

Distributiv funktsiyasidan foydalangan holda 3d+2 umumiy terminini chiqaring.

$$\left(3d+2\right)\left(d+6\right)$$

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.

$$3d^{2}+20d+12=0$$

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.

$$d=\frac{-20±\sqrt{20^{2}-4\times 3\times 12}}{2\times 3}$$

20 kvadratini chiqarish.

$$d=\frac{-20±\sqrt{400-4\times 3\times 12}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$d=\frac{-20±\sqrt{400-12\times 12}}{2\times 3}$$

-12 ni 12 marotabaga ko'paytirish.

$$d=\frac{-20±\sqrt{400-144}}{2\times 3}$$

400 ni -144 ga qo'shish.

$$d=\frac{-20±\sqrt{256}}{2\times 3}$$

256 ning kvadrat ildizini chiqarish.

$$d=\frac{-20±16}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$d=\frac{-20±16}{6}$$

d=\frac{-20±16}{6} tenglamasini yeching, bunda ± musbat. -20 ni 16 ga qo'shish.

$$d=-\frac{4}{6}$$

\frac{-4}{6} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

d=\frac{-20±16}{6} tenglamasini yeching, bunda ± manfiy. -20 dan 16 ni ayirish.

$$d=-\frac{36}{6}$$

-36 ni 6 ga bo'lish.

$$d=-6$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\frac{2}{3} ga va x_{2} uchun -6 ga bo‘ling.

$$3d^{2}+20d+12=3\left(d-\left(-\frac{2}{3}\right)\right)\left(d-\left(-6\right)\right)$$

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

$$3d^{2}+20d+12=3\left(d+\frac{2}{3}\right)\left(d+6\right)$$

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

$$3d^{2}+20d+12=3\times \frac{3d+2}{3}\left(d+6\right)$$

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

$$3d^{2}+20d+12=\left(3d+2\right)\left(d+6\right)$$

Baholash

$$\left(d+6\right)\left(3d+2\right)$$