Question

Example 6: Divide (6 - 13x + 6x?) by (3 + 2x)

only little Moon will ans ​

Answer 1

Answer:

$\huge\mathfrak\red{solutions}$

$\tt\pink{step1}$ => write the divisior (-3+2x) and dividend (6-13x+6x²) any standard form.

dividend : 6x² - 13x + 6

divisor : 2x - 3

$\tt\pink{step2}$ => divide the first term of the dividend (6x²) by the first term of the divisor (2x) to get the first term of the quotient (3x) (i.e. 6x²÷2x = 3x)

$\tt\pink{step3}$ => multiply divisior (2 x - 3 ) by the first term of the quotient (3x) . subtract the product of (2x-3) and (3x) , i.e. [3x (2x - 3) => 6x² - 9x]

from the dividend (6x² - 13x + 6) to get the remainder

[6x²-13x+6-(6x²-9x)=-4x+6].

$\tt\pink{step4}$ => considered the reminder (-4x+6) as the new dividend but the divisior will be the same . divide the first term of the dividend (-4x) by the first term of the day after (2x) to get the second term of the quotient (i.e. -4x÷2x=-2).

$\tt\pink{step5}$ => multiply the divisior (2 x-3) by -2 , i.e. the second term of the quotient subtract the product -2(2x-3) from (-4x+6)

i.e. -4x+6-[-2(2x-3)] = -4x +6+4x-6 = 0

the reminder is zero .

hence , (6x² - 13x + 6) ÷ (2x-3) => 3x - 2

Answer 2

Answer:

solutions

\tt\pink{step1}step1 => write the divisior (-3+2x) and dividend (6-13x+6x²) any standard form.

dividend : 6x² - 13x + 6

divisor : 2x - 3

\tt\pink{step2}step2 => divide the first term of the dividend (6x²) by the first term of the divisor (2x) to get the first term of the quotient (3x) (i.e. 6x²÷2x = 3x)

\tt\pink{step3}step3 => multiply divisior (2 x - 3 ) by the first term of the quotient (3x) . subtract the product of (2x-3) and (3x) , i.e. [3x (2x - 3) => 6x² - 9x]

from the dividend (6x² - 13x + 6) to get the remainder

[6x²-13x+6-(6x²-9x)=-4x+6].

\tt\pink{step4}step4 => considered the reminder (-4x+6) as the new dividend but the divisior will be the same . divide the first term of the dividend (-4x) by the first term of the day after (2x) to get the second term of the quotient (i.e. -4x÷2x=-2).

\tt\pink{step5}step5 => multiply the divisior (2 x-3) by -2 , i.e. the second term of the quotient subtract the product -2(2x-3) from (-4x+6)

i.e. -4x+6-[-2(2x-3)] = -4x +6+4x-6 = 0

the reminder is zero .

hence , (6x² - 13x + 6) ÷ (2x-3) => 3x - 2