Question

How do you multiply `\frac { 2x ^ { 2} - 4x + 8} { 6x ^ { 2} + 7x - 3} \cdot \frac { 2x ^ { 2} + x - 3} { x ^ { 3} + 8}`?

Answer 1

`[2(x-1)]/[(x+2)(3x-1)]`

Explanation:

Given `[2x^2-4x+8]/[6x^2+7x-3]. [2x^2+x-3]/[x^3+8]`

`rArr [2(x^2-2x+4)]/[6x^2+(9-2)x-3].[2x^2+(3-2)x-3]/[x^3+2^3]`

`rArr [2(x^2-2x+4)]/[6x^2+9x-2x-3].[2x^2+3x-2x-3]/[(x+2)(x^2-2x+4)]`

`rArr [2 {cancel(x^2-2x+4)}]/[3x(2x+3)-1(2x+3)].[x(2x+3)-1(2x+3)]/[(x+2){cancel(x^2-2x+4)]`

`rArr [2]/[(3x-1)(2x+3)].[(x-1)(2x+3)]/[(x+2)]`

`rArr [2(x-1){cancel(2x+3)}]/[(x+2)(3x-1){cancel(2x+3)}]`

`rArr [2(x-1)]/[(x+2)(3x-1)]`