f(x)=5x^2+10x-15 eching

$$f ( x ) = 5 x ^ { 2 } + 10 x - 15$$

Omil

$$5\left(x-1\right)\left(x+3\right)$$

Yechim qadamlari

5 omili.

$$5\left(x^{2}+2x-3\right)$$

Hisoblang: x^{2}+2x-3. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda x^{2}+ax+bx-3 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=2 ab=1\left(-3\right)=-3$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-1 b=3$$

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

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

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

$$x\left(x-1\right)+3\left(x-1\right)$$

Distributiv funktsiyasidan foydalangan holda x-1 umumiy terminini chiqaring.

$$\left(x-1\right)\left(x+3\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$5\left(x-1\right)\left(x+3\right)$$

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.

$$5x^{2}+10x-15=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5\left(-15\right)}}{2\times 5}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 5\left(-15\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-20\left(-15\right)}}{2\times 5}$$

-20 ni -15 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100+300}}{2\times 5}$$

100 ni 300 ga qo'shish.

$$x=\frac{-10±\sqrt{400}}{2\times 5}$$

400 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±20}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±20}{10}$$

x=\frac{-10±20}{10} tenglamasini yeching, bunda ± musbat. -10 ni 20 ga qo'shish.

$$x=\frac{10}{10}$$

10 ni 10 ga bo'lish.

$$x=1$$

x=\frac{-10±20}{10} tenglamasini yeching, bunda ± manfiy. -10 dan 20 ni ayirish.

$$x=-\frac{30}{10}$$

-30 ni 10 ga bo'lish.

$$x=-3$$

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.

$$5x^{2}+10x-15=5\left(x-1\right)\left(x-\left(-3\right)\right)$$

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

$$5x^{2}+10x-15=5\left(x-1\right)\left(x+3\right)$$

Baholash

$$5\left(x-1\right)\left(x+3\right)$$