What is (2x^3+10x^2+16x+8)/(4x^3+8x^2-4x-8) ?

The solution to (2x^3+10x^2+16x+8)/(4x^3+8x^2-4x-8) is (x+2)/(2(x-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (2x^3+10x^2+16x+8)/(4x^3+8x^2-4x-8)

simplify

\frac{2 x ^{3} + 10 x ^{2} + 16 x + 8}{4 x ^{3} + 8 x ^{2} - 4 x - 8}

\frac{x + 2}{2 ( x - 1 )}

Solution steps

\frac{2 x ^{3} + 10 x ^{2} + 16 x + 8}{4 x ^{3} + 8 x ^{2} - 4 x - 8}

Factor

2 x^{3} + 10 x^{2} + 16 x + 8 : 2 (x + 1) (x + 2)^{2}

= \frac{2 ( x + 1 ) ( x + 2 ) ^{2}}{4 x ^{3} + 8 x ^{2} - 4 x - 8}

Factor

4 x^{3} + 8 x^{2} - 4 x - 8 : 4 (x + 2) (x + 1) (x - 1)

= \frac{2 ( x + 1 ) ( x + 2 ) ^{2}}{4 ( x + 2 ) ( x + 1 ) ( x - 1 )}

Cancel the common factor:

x + 1

= \frac{2 ( x + 2 ) ^{2}}{4 ( x + 2 ) ( x - 1 )}

Factor the number:

4 = 2 \cdot 2

= \frac{2 ( x + 2 ) ^{2}}{2 \cdot 2 ( x + 2 ) ( x - 1 )}

Cancel the common factor:

2

= \frac{( x + 2 ) ^{2}}{2 ( x + 2 ) ( x - 1 )}

Apply exponent rule:

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

(x + 2)^{2} = (x + 2) (x + 2)

= \frac{( x + 2 ) ( x + 2 )}{2 ( x + 2 ) ( x - 1 )}

Cancel the common factor:

x + 2

= \frac{x + 2}{2 ( x - 1 )}

(2 x - 5) (4 x^{3} - 7 x^{2} + 2 x - 3)

ln (\frac{1}{e ^{11}})

4 - 4 x^{2}

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

\frac{1}{2} \div \frac{3}{8}

What is (2x^3+10x^2+16x+8)/(4x^3+8x^2-4x-8) ?
The solution to (2x^3+10x^2+16x+8)/(4x^3+8x^2-4x-8) is (x+2)/(2(x-1))