b^2+4b+3 eching

$$b ^ { 2 } + 4 b + 3$$

Omil

$$\left(b+1\right)\left(b+3\right)$$

Yechim qadamlari

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

$$p+q=4 pq=1\times 3=3$$

pq musbat boʻlganda, p va q da bir xil belgi bor. p+q musbat boʻlganda, p va q ikkisi ham musbat. Faqat bundan juftlik tizim yechimidir.

$$p=1 q=3$$

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

$$\left(b^{2}+b\right)+\left(3b+3\right)$$

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

$$b\left(b+1\right)+3\left(b+1\right)$$

Distributiv funktsiyasidan foydalangan holda b+1 umumiy terminini chiqaring.

$$\left(b+1\right)\left(b+3\right)$$

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.

$$b^{2}+4b+3=0$$

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.

$$b=\frac{-4±\sqrt{4^{2}-4\times 3}}{2}$$

4 kvadratini chiqarish.

$$b=\frac{-4±\sqrt{16-4\times 3}}{2}$$

-4 ni 3 marotabaga ko'paytirish.

$$b=\frac{-4±\sqrt{16-12}}{2}$$

16 ni -12 ga qo'shish.

$$b=\frac{-4±\sqrt{4}}{2}$$

4 ning kvadrat ildizini chiqarish.

$$b=\frac{-4±2}{2}$$

b=\frac{-4±2}{2} tenglamasini yeching, bunda ± musbat. -4 ni 2 ga qo'shish.

$$b=-\frac{2}{2}$$

-2 ni 2 ga bo'lish.

$$b=-1$$

b=\frac{-4±2}{2} tenglamasini yeching, bunda ± manfiy. -4 dan 2 ni ayirish.

$$b=-\frac{6}{2}$$

-6 ni 2 ga bo'lish.

$$b=-3$$

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

$$b^{2}+4b+3=\left(b-\left(-1\right)\right)\left(b-\left(-3\right)\right)$$

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

$$b^{2}+4b+3=\left(b+1\right)\left(b+3\right)$$

Baholash

$$\left(b+1\right)\left(b+3\right)$$