3p^2-3p-36 eching

$$3 p ^ { 2 } - 3 p - 36$$

Omil

$$3\left(p-4\right)\left(p+3\right)$$

Yechim qadamlari

3 omili.

$$3\left(p^{2}-p-12\right)$$

Hisoblang: p^{2}-p-12. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda p^{2}+ap+bp-12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=1\left(-12\right)=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -12-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-12 2,-6 3,-4$$

Har bir juftlik yigʻindisini hisoblang.

$$1-12=-11 2-6=-4 3-4=-1$$

Yechim – -1 yigʻindisini beruvchi juftlik.

$$a=-4 b=3$$

p^{2}-p-12 ni \left(p^{2}-4p\right)+\left(3p-12\right) sifatida qaytadan yozish.

$$\left(p^{2}-4p\right)+\left(3p-12\right)$$

Birinchi guruhda p ni va ikkinchi guruhda 3 ni faktordan chiqaring.

$$p\left(p-4\right)+3\left(p-4\right)$$

Distributiv funktsiyasidan foydalangan holda p-4 umumiy terminini chiqaring.

$$\left(p-4\right)\left(p+3\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$3\left(p-4\right)\left(p+3\right)$$

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.

$$3p^{2}-3p-36=0$$

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.

$$p=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-36\right)}}{2\times 3}$$

-3 kvadratini chiqarish.

$$p=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-36\right)}}{2\times 3}$$

-4 ni 3 marotabaga ko'paytirish.

$$p=\frac{-\left(-3\right)±\sqrt{9-12\left(-36\right)}}{2\times 3}$$

-12 ni -36 marotabaga ko'paytirish.

$$p=\frac{-\left(-3\right)±\sqrt{9+432}}{2\times 3}$$

9 ni 432 ga qo'shish.

$$p=\frac{-\left(-3\right)±\sqrt{441}}{2\times 3}$$

441 ning kvadrat ildizini chiqarish.

$$p=\frac{-\left(-3\right)±21}{2\times 3}$$

-3 ning teskarisi 3 ga teng.

$$p=\frac{3±21}{2\times 3}$$

2 ni 3 marotabaga ko'paytirish.

$$p=\frac{3±21}{6}$$

p=\frac{3±21}{6} tenglamasini yeching, bunda ± musbat. 3 ni 21 ga qo'shish.

$$p=\frac{24}{6}$$

24 ni 6 ga bo'lish.

$$p=4$$

p=\frac{3±21}{6} tenglamasini yeching, bunda ± manfiy. 3 dan 21 ni ayirish.

$$p=-\frac{18}{6}$$

-18 ni 6 ga bo'lish.

$$p=-3$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 4 ga va x_{2} uchun -3 ga bo‘ling.

$$3p^{2}-3p-36=3\left(p-4\right)\left(p-\left(-3\right)\right)$$

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

$$3p^{2}-3p-36=3\left(p-4\right)\left(p+3\right)$$

Baholash

$$3\left(p-4\right)\left(p+3\right)$$