How do you evaluate \frac { 2x ^ { 2} + 2x } { 2x ^ { 2} } \cdot \frac { x ^ { 2} - 3x } { x ^ { 2} - 2x - 3}?

2 Answers
Jan 14, 2018

1

Explanation:

(2x^2+2x)/(2x^2)*(x^2-3x)/(x^2-2x-3)

=(2x(x+1))/(2x^2)*(x(x-3))/((x+1)(x-3))

=((cancel(2x)cancel((x+1)))/cancel(2x^2)*(cancel(x)cancel((x-3)))/(cancel((x+1))cancel((x-3))))

=1

Jan 14, 2018

1

Explanation:

"factorise and cancel common factors where possible"

2x^2+2x=2x(x+1)larrcolor(blue)"common factor of 2x"

x^2-3x=x(x-3)larrcolor(blue)"common factor of x"

x2-2x-3=(x-3)(x+1)larrcolor(blue)"quadratic factorising"

rArr(2x^2+2x)/(2x^2)xx(x^2-3x)/(x^2-2x-3)

=(cancel(2x)cancel((x+1)))/(cancel(2x)cancel(x))xx(cancel(x)cancel((x-3)))/(cancel((x-3))cancel((x+1)))

=1xx1=1