5x^2-16x-21 eching

$$5 { x }^{ 2 } -16x-21$$

Omil

$$\left(5x-21\right)\left(x+1\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 5x^{2}+ax+bx-21 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-16 ab=5\left(-21\right)=-105$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -105-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-105 3,-35 5,-21 7,-15$$

Har bir juftlik yigʻindisini hisoblang.

$$1-105=-104 3-35=-32 5-21=-16 7-15=-8$$

Yechim – -16 yigʻindisini beruvchi juftlik.

$$a=-21 b=5$$

5x^{2}-16x-21 ni \left(5x^{2}-21x\right)+\left(5x-21\right) sifatida qaytadan yozish.

$$\left(5x^{2}-21x\right)+\left(5x-21\right)$$

5x^{2}-21x ichida x ni ajrating.

$$x\left(5x-21\right)+5x-21$$

Distributiv funktsiyasidan foydalangan holda 5x-21 umumiy terminini chiqaring.

$$\left(5x-21\right)\left(x+1\right)$$

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.

$$5x^{2}-16x-21=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\left(-21\right)}}{2\times 5}$$

-16 kvadratini chiqarish.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 5\left(-21\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-\left(-16\right)±\sqrt{256-20\left(-21\right)}}{2\times 5}$$

-20 ni -21 marotabaga ko'paytirish.

$$x=\frac{-\left(-16\right)±\sqrt{256+420}}{2\times 5}$$

256 ni 420 ga qo'shish.

$$x=\frac{-\left(-16\right)±\sqrt{676}}{2\times 5}$$

676 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-16\right)±26}{2\times 5}$$

-16 ning teskarisi 16 ga teng.

$$x=\frac{16±26}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{16±26}{10}$$

x=\frac{16±26}{10} tenglamasini yeching, bunda ± musbat. 16 ni 26 ga qo'shish.

$$x=\frac{42}{10}$$

\frac{42}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{21}{5}$$

x=\frac{16±26}{10} tenglamasini yeching, bunda ± manfiy. 16 dan 26 ni ayirish.

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

-10 ni 10 ga bo'lish.

$$x=-1$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{21}{5} ga va x_{2} uchun -1 ga bo‘ling.

$$5x^{2}-16x-21=5\left(x-\frac{21}{5}\right)\left(x-\left(-1\right)\right)$$

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

$$5x^{2}-16x-21=5\left(x-\frac{21}{5}\right)\left(x+1\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{21}{5} ni x dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$5x^{2}-16x-21=5\times \frac{5x-21}{5}\left(x+1\right)$$

5 va 5 ichida eng katta umumiy 5 faktorini bekor qiling.

$$5x^{2}-16x-21=\left(5x-21\right)\left(x+1\right)$$

Baholash

$$\left(5x-21\right)\left(x+1\right)$$