What is (4ab^3-3a^2bc+12a^3b^2c^4)/(-2ab^2c^3) ?

The solution to (4ab^3-3a^2bc+12a^3b^2c^4)/(-2ab^2c^3) is -(4b^2-3ac+12a^2bc^4)/(2bc^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (4ab^3-3a^2bc+12a^3b^2c^4)/(-2ab^2c^3)

simplify

\frac{4 a b ^{3} - 3 a ^{2} b c + 12 a ^{3} b ^{2} c ^{4}}{- 2 a b ^{2} c ^{3}}

- \frac{4 b ^{2} - 3 a c + 12 a ^{2} b c ^{4}}{2 b c ^{3}}

Solution steps

\frac{4 a b ^{3} - 3 a ^{2} b c + 12 a ^{3} b ^{2} c ^{4}}{- 2 a b ^{2} c ^{3}}

Factor out common term

ab : ab (4 b^{2} - 3 a c + 12 a^{2} b c^{4})

= \frac{ab ( 4 b ^{2} - 3 a c + 12 a ^{2} b c ^{4} )}{- 2 a b ^{2} c ^{3}}

Cancel the common factor:

a

= \frac{b ( 4 b ^{2} - 3 a c + 12 a ^{2} b c ^{4} )}{- 2 b ^{2} c ^{3}}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

b^{2} = bb

= \frac{b ( 4 b ^{2} - 3 a c + 12 a ^{2} b c ^{4} )}{- 2 bb c ^{3}}

Cancel the common factor:

b

= \frac{4 b ^{2} - 3 a c + 12 a ^{2} b c ^{4}}{- 2 b c ^{3}}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{4 b ^{2} - 3 a c + 12 a ^{2} b c ^{4}}{2 b c ^{3}}

(3 - 1)^{2} + 2 (3 - 1) + 5

(x^{2} - 8)^{2} - 8

\frac{2 5}{3}

(2 k + 1)^{4}

\frac{7}{2 x ^{2} + 18 x} - \frac{5 x}{x ^{2} + 17 x + 72}

What is (4ab^3-3a^2bc+12a^3b^2c^4)/(-2ab^2c^3) ?
The solution to (4ab^3-3a^2bc+12a^3b^2c^4)/(-2ab^2c^3) is -(4b^2-3ac+12a^2bc^4)/(2bc^3)