4t^2+12t eching

$$4 t ^ { 2 } + 12 t$$

Omil

$$4t\left(t+3\right)$$

Yechim qadamlari

4 omili.

$$4\left(t^{2}+3t\right)$$

Hisoblang: t^{2}+3t. t omili.

$$t\left(t+3\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$4t\left(t+3\right)$$

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.

$$4t^{2}+12t=0$$

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.

$$t=\frac{-12±\sqrt{12^{2}}}{2\times 4}$$

12^{2} ning kvadrat ildizini chiqarish.

$$t=\frac{-12±12}{2\times 4}$$

2 ni 4 marotabaga ko'paytirish.

$$t=\frac{-12±12}{8}$$

t=\frac{-12±12}{8} tenglamasini yeching, bunda ± musbat. -12 ni 12 ga qo'shish.

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

0 ni 8 ga bo'lish.

$$t=0$$

t=\frac{-12±12}{8} tenglamasini yeching, bunda ± manfiy. -12 dan 12 ni ayirish.

$$t=-\frac{24}{8}$$

-24 ni 8 ga bo'lish.

$$t=-3$$

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 -3 ga bo‘ling.

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

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

$$4t^{2}+12t=4t\left(t+3\right)$$

Baholash

$$4t\left(t+3\right)$$