5z^2-33z+18 eching

$$5 z ^ { 2 } - 33 z + 18$$

Omil

$$\left(z-6\right)\left(5z-3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 5z^{2}+az+bz+18 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-33 ab=5\times 18=90$$

ab musbat boʻlganda, a va b da bir xil belgi bor. a+b manfiy boʻlganda, a va b ikkisi ham manfiy. 90-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-90 -2,-45 -3,-30 -5,-18 -6,-15 -9,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-90=-91 -2-45=-47 -3-30=-33 -5-18=-23 -6-15=-21 -9-10=-19$$

Yechim – -33 yigʻindisini beruvchi juftlik.

$$a=-30 b=-3$$

5z^{2}-33z+18 ni \left(5z^{2}-30z\right)+\left(-3z+18\right) sifatida qaytadan yozish.

$$\left(5z^{2}-30z\right)+\left(-3z+18\right)$$

Birinchi guruhda 5z ni va ikkinchi guruhda -3 ni faktordan chiqaring.

$$5z\left(z-6\right)-3\left(z-6\right)$$

Distributiv funktsiyasidan foydalangan holda z-6 umumiy terminini chiqaring.

$$\left(z-6\right)\left(5z-3\right)$$

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.

$$5z^{2}-33z+18=0$$

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.

$$z=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 5\times 18}}{2\times 5}$$

-33 kvadratini chiqarish.

$$z=\frac{-\left(-33\right)±\sqrt{1089-4\times 5\times 18}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$z=\frac{-\left(-33\right)±\sqrt{1089-20\times 18}}{2\times 5}$$

-20 ni 18 marotabaga ko'paytirish.

$$z=\frac{-\left(-33\right)±\sqrt{1089-360}}{2\times 5}$$

1089 ni -360 ga qo'shish.

$$z=\frac{-\left(-33\right)±\sqrt{729}}{2\times 5}$$

729 ning kvadrat ildizini chiqarish.

$$z=\frac{-\left(-33\right)±27}{2\times 5}$$

-33 ning teskarisi 33 ga teng.

$$z=\frac{33±27}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$z=\frac{33±27}{10}$$

z=\frac{33±27}{10} tenglamasini yeching, bunda ± musbat. 33 ni 27 ga qo'shish.

$$z=\frac{60}{10}$$

60 ni 10 ga bo'lish.

$$z=6$$

z=\frac{33±27}{10} tenglamasini yeching, bunda ± manfiy. 33 dan 27 ni ayirish.

$$z=\frac{6}{10}$$

\frac{6}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$z=\frac{3}{5}$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 6 ga va x_{2} uchun \frac{3}{5} ga bo‘ling.

$$5z^{2}-33z+18=5\left(z-6\right)\left(z-\frac{3}{5}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{3}{5} ni z dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$5z^{2}-33z+18=5\left(z-6\right)\times \frac{5z-3}{5}$$

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

$$5z^{2}-33z+18=\left(z-6\right)\left(5z-3\right)$$

Baholash

$$\left(z-6\right)\left(5z-3\right)$$