=25a^2-35a+12 eching

$$= 25 a ^ { 2 } - 35 a + 12$$

Omil

$$\left(5a-4\right)\left(5a-3\right)$$

Yechim qadamlari

Ifodani guruhlash orqali faktorlang. Avvalo, ifoda 25a^{2}+pa+qa+12 sifatida qayta yozilishi kerak. p va q ni topish uchun yechiladigan tizimni sozlang.

$$p+q=-35 pq=25\times 12=300$$

pq musbat boʻlganda, p va q da bir xil belgi bor. p+q manfiy boʻlganda, p va q ikkisi ham manfiy. 300-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$-1,-300 -2,-150 -3,-100 -4,-75 -5,-60 -6,-50 -10,-30 -12,-25 -15,-20$$

Har bir juftlik yigʻindisini hisoblang.

$$-1-300=-301 -2-150=-152 -3-100=-103 -4-75=-79 -5-60=-65 -6-50=-56 -10-30=-40 -12-25=-37 -15-20=-35$$

Yechim – -35 yigʻindisini beruvchi juftlik.

$$p=-20 q=-15$$

25a^{2}-35a+12 ni \left(25a^{2}-20a\right)+\left(-15a+12\right) sifatida qaytadan yozish.

$$\left(25a^{2}-20a\right)+\left(-15a+12\right)$$

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

$$5a\left(5a-4\right)-3\left(5a-4\right)$$

Distributiv funktsiyasidan foydalangan holda 5a-4 umumiy terminini chiqaring.

$$\left(5a-4\right)\left(5a-3\right)$$

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.

$$25a^{2}-35a+12=0$$

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.

$$a=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 25\times 12}}{2\times 25}$$

-35 kvadratini chiqarish.

$$a=\frac{-\left(-35\right)±\sqrt{1225-4\times 25\times 12}}{2\times 25}$$

-4 ni 25 marotabaga ko'paytirish.

$$a=\frac{-\left(-35\right)±\sqrt{1225-100\times 12}}{2\times 25}$$

-100 ni 12 marotabaga ko'paytirish.

$$a=\frac{-\left(-35\right)±\sqrt{1225-1200}}{2\times 25}$$

1225 ni -1200 ga qo'shish.

$$a=\frac{-\left(-35\right)±\sqrt{25}}{2\times 25}$$

25 ning kvadrat ildizini chiqarish.

$$a=\frac{-\left(-35\right)±5}{2\times 25}$$

-35 ning teskarisi 35 ga teng.

$$a=\frac{35±5}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$a=\frac{35±5}{50}$$

a=\frac{35±5}{50} tenglamasini yeching, bunda ± musbat. 35 ni 5 ga qo'shish.

$$a=\frac{40}{50}$$

\frac{40}{50} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$a=\frac{4}{5}$$

a=\frac{35±5}{50} tenglamasini yeching, bunda ± manfiy. 35 dan 5 ni ayirish.

$$a=\frac{30}{50}$$

\frac{30}{50} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun \frac{4}{5} ga va x_{2} uchun \frac{3}{5} ga bo‘ling.

$$25a^{2}-35a+12=25\left(a-\frac{4}{5}\right)\left(a-\frac{3}{5}\right)$$

Umumiy maxrajni topib va suratlarni ayirib \frac{4}{5} ni a dan ayirish. So'ngra imkoni boricha kasrni eng kichik shartga qisqartirish.

$$25a^{2}-35a+12=25\times \frac{5a-4}{5}\left(a-\frac{3}{5}\right)$$

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

$$25a^{2}-35a+12=25\times \frac{5a-4}{5}\times \frac{5a-3}{5}$$

Raqamlash sonlarini va maxraj sonlariga ko'paytirish orqali \frac{5a-4}{5} ni \frac{5a-3}{5} ga ko'paytirish. So'ngra kasrni imkoni boricha eng kam a'zoga qisqartiring.

$$25a^{2}-35a+12=25\times \frac{\left(5a-4\right)\left(5a-3\right)}{5\times 5}$$

5 ni 5 marotabaga ko'paytirish.

$$25a^{2}-35a+12=25\times \frac{\left(5a-4\right)\left(5a-3\right)}{25}$$

25 va 25 ichida eng katta umumiy 25 faktorini bekor qiling.

$$25a^{2}-35a+12=\left(5a-4\right)\left(5a-3\right)$$

Baholash

$$\left(5a-4\right)\left(5a-3\right)$$