-8x^2-9x+17 eching

$$- 8 x ^ { 2 } - 9 x + 17$$

Omil

$$\left(1-x\right)\left(8x+17\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda -8x^{2}+ax+bx+17 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-9 ab=-8\times 17=-136$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -136-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-136 2,-68 4,-34 8,-17$$

Har bir juftlik yigʻindisini hisoblang.

$$1-136=-135 2-68=-66 4-34=-30 8-17=-9$$

Yechim – -9 yigʻindisini beruvchi juftlik.

$$a=8 b=-17$$

-8x^{2}-9x+17 ni \left(-8x^{2}+8x\right)+\left(-17x+17\right) sifatida qaytadan yozish.

$$\left(-8x^{2}+8x\right)+\left(-17x+17\right)$$

Birinchi guruhda 8x ni va ikkinchi guruhda 17 ni faktordan chiqaring.

$$8x\left(-x+1\right)+17\left(-x+1\right)$$

Distributiv funktsiyasidan foydalangan holda -x+1 umumiy terminini chiqaring.

$$\left(-x+1\right)\left(8x+17\right)$$

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.

$$-8x^{2}-9x+17=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-8\right)\times 17}}{2\left(-8\right)}$$

-9 kvadratini chiqarish.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-8\right)\times 17}}{2\left(-8\right)}$$

-4 ni -8 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81+32\times 17}}{2\left(-8\right)}$$

32 ni 17 marotabaga ko'paytirish.

$$x=\frac{-\left(-9\right)±\sqrt{81+544}}{2\left(-8\right)}$$

81 ni 544 ga qo'shish.

$$x=\frac{-\left(-9\right)±\sqrt{625}}{2\left(-8\right)}$$

625 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-9\right)±25}{2\left(-8\right)}$$

-9 ning teskarisi 9 ga teng.

$$x=\frac{9±25}{2\left(-8\right)}$$

2 ni -8 marotabaga ko'paytirish.

$$x=\frac{9±25}{-16}$$

x=\frac{9±25}{-16} tenglamasini yeching, bunda ± musbat. 9 ni 25 ga qo'shish.

$$x=\frac{34}{-16}$$

\frac{34}{-16} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{17}{8}$$

x=\frac{9±25}{-16} tenglamasini yeching, bunda ± manfiy. 9 dan 25 ni ayirish.

$$x=-\frac{16}{-16}$$

-16 ni -16 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{17}{8} ga va x_{2} uchun 1 ga bo‘ling.

$$-8x^{2}-9x+17=-8\left(x-\left(-\frac{17}{8}\right)\right)\left(x-1\right)$$

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

$$-8x^{2}-9x+17=-8\left(x+\frac{17}{8}\right)\left(x-1\right)$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{17}{8} ni x ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$-8x^{2}-9x+17=-8\times \frac{-8x-17}{-8}\left(x-1\right)$$

-8 va 8 ichida eng katta umumiy 8 faktorini bekor qiling.

$$-8x^{2}-9x+17=\left(-8x-17\right)\left(x-1\right)$$

Baholash

$$\left(1-x\right)\left(8x+17\right)$$