h(x)=-5x^2+20x+60 eching

$$h ( x ) = - 5 x ^ { 2 } + 20 x + 60$$

Omil

$$-5\left(x-6\right)\left(x+2\right)$$

Yechim qadamlari

5 omili.

$$5\left(-x^{2}+4x+12\right)$$

Hisoblang: -x^{2}+4x+12. Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -x^{2}+ax+bx+12 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=4 ab=-12=-12$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b musbat boʻlganda, musbat sonda manfiyga 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 – 4 yigʻindisini beruvchi juftlik.

$$a=6 b=-2$$

-x^{2}+4x+12 ni \left(-x^{2}+6x\right)+\left(-2x+12\right) sifatida qaytadan yozish.

$$\left(-x^{2}+6x\right)+\left(-2x+12\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -2 ni faktordan chiqaring.

$$-x\left(x-6\right)-2\left(x-6\right)$$

Distributiv funktsiyasidan foydalangan holda x-6 umumiy terminini chiqaring.

$$\left(x-6\right)\left(-x-2\right)$$

Toʻliq ajratilgan ifodani qaytadan yozing.

$$5\left(x-6\right)\left(-x-2\right)$$

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.

$$-5x^{2}+20x+60=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-5\right)\times 60}}{2\left(-5\right)}$$

20 kvadratini chiqarish.

$$x=\frac{-20±\sqrt{400-4\left(-5\right)\times 60}}{2\left(-5\right)}$$

-4 ni -5 marotabaga ko'paytirish.

$$x=\frac{-20±\sqrt{400+20\times 60}}{2\left(-5\right)}$$

20 ni 60 marotabaga ko'paytirish.

$$x=\frac{-20±\sqrt{400+1200}}{2\left(-5\right)}$$

400 ni 1200 ga qo'shish.

$$x=\frac{-20±\sqrt{1600}}{2\left(-5\right)}$$

1600 ning kvadrat ildizini chiqarish.

$$x=\frac{-20±40}{2\left(-5\right)}$$

2 ni -5 marotabaga ko'paytirish.

$$x=\frac{-20±40}{-10}$$

x=\frac{-20±40}{-10} tenglamasini yeching, bunda ± musbat. -20 ni 40 ga qo'shish.

$$x=\frac{20}{-10}$$

20 ni -10 ga bo'lish.

$$x=-2$$

x=\frac{-20±40}{-10} tenglamasini yeching, bunda ± manfiy. -20 dan 40 ni ayirish.

$$x=-\frac{60}{-10}$$

-60 ni -10 ga bo'lish.

$$x=6$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -2 ga va x_{2} uchun 6 ga bo‘ling.

$$-5x^{2}+20x+60=-5\left(x-\left(-2\right)\right)\left(x-6\right)$$

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

$$-5x^{2}+20x+60=-5\left(x+2\right)\left(x-6\right)$$

Baholash

$$-5\left(x-6\right)\left(x+2\right)$$