4a^2+8a eching

$$4 a ^ { 2 } + 8 a$$

Omil

$$4a\left(a+2\right)$$

Yechim qadamlari

4 omili.

$$4\left(a^{2}+2a\right)$$

Hisoblang: a^{2}+2a. a omili.

$$a\left(a+2\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$4a\left(a+2\right)$$

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.

$$4a^{2}+8a=0$$

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.

$$a=\frac{-8±\sqrt{8^{2}}}{2\times 4}$$

8^{2} ning kvadrat ildizini chiqarish.

$$a=\frac{-8±8}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$a=\frac{-8±8}{8}$$

a=\frac{-8±8}{8} tenglamasini yeching, bunda ± musbat. -8 ni 8 ga qo'shish.

$$a=\frac{0}{8}$$

0 ni 8 ga bo'lish.

$$a=0$$

a=\frac{-8±8}{8} tenglamasini yeching, bunda ± manfiy. -8 dan 8 ni ayirish.

$$a=-\frac{16}{8}$$

-16 ni 8 ga bo'lish.

$$a=-2$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 0 ga va x_{2} uchun -2 ga bo‘ling.

$$4a^{2}+8a=4a\left(a-\left(-2\right)\right)$$

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

$$4a^{2}+8a=4a\left(a+2\right)$$

Baholash

$$4a\left(a+2\right)$$