2z^2-23z+30 eching

$$2 z ^ { 2 } - 23 z + 30$$

Omil

$$\left(z-10\right)\left(2z-3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 2z^{2}+az+bz+30 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-23 ab=2\times 30=60$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 60-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

Yechim – -23 yigʻindisini beruvchi juftlik.

$$a=-20 b=-3$$

2z^{2}-23z+30 ni \left(2z^{2}-20z\right)+\left(-3z+30\right) sifatida qaytadan yozish.

$$\left(2z^{2}-20z\right)+\left(-3z+30\right)$$

Birinchi guruhda 2z ni va ikkinchi guruhda -3 ni faktordan chiqaring.

$$2z\left(z-10\right)-3\left(z-10\right)$$

Distributiv funktsiyasidan foydalangan holda z-10 umumiy terminini chiqaring.

$$\left(z-10\right)\left(2z-3\right)$$

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.

$$2z^{2}-23z+30=0$$

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.

$$z=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 2\times 30}}{2\times 2}$$

-23 kvadratini chiqarish.

$$z=\frac{-\left(-23\right)±\sqrt{529-4\times 2\times 30}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$z=\frac{-\left(-23\right)±\sqrt{529-8\times 30}}{2\times 2}$$

-8 ni 30 marotabaga ko'paytirish.

$$z=\frac{-\left(-23\right)±\sqrt{529-240}}{2\times 2}$$

529 ni -240 ga qo'shish.

$$z=\frac{-\left(-23\right)±\sqrt{289}}{2\times 2}$$

289 ning kvadrat ildizini chiqarish.

$$z=\frac{-\left(-23\right)±17}{2\times 2}$$

-23 ning teskarisi 23 ga teng.

$$z=\frac{23±17}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$z=\frac{23±17}{4}$$

z=\frac{23±17}{4} tenglamasini yeching, bunda ± musbat. 23 ni 17 ga qo'shish.

$$z=\frac{40}{4}$$

40 ni 4 ga bo'lish.

$$z=10$$

z=\frac{23±17}{4} tenglamasini yeching, bunda ± manfiy. 23 dan 17 ni ayirish.

$$z=\frac{6}{4}$$

\frac{6}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$z=\frac{3}{2}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 10 ga va x_{2} uchun \frac{3}{2} ga bo‘ling.

$$2z^{2}-23z+30=2\left(z-10\right)\left(z-\frac{3}{2}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{2} ni z dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$2z^{2}-23z+30=2\left(z-10\right)\times \frac{2z-3}{2}$$

2 va 2 ichida eng katta umumiy 2 faktorini bekor qiling.

$$2z^{2}-23z+30=\left(z-10\right)\left(2z-3\right)$$

Baholash

$$\left(z-10\right)\left(2z-3\right)$$