Question

a) (x^2-2x+3).(1/2x-5) b) (x^2y^2 -1/3xy+2y).(x-2y)

Answer 1

\(a,\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\\ =x^2.\dfrac{1}{2}x-5x^2-2x.\dfrac{1}{2}x+2x.5+3.\dfrac{1}{2}x-15\\ =\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\\ =\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

\(b,\left(x^2y^2-\dfrac{1}{3}xy+2y\right)\left(x-2y\right)\\ =x^3y-2x^2y^3-\dfrac{1}{3}x^2y+\dfrac{2}{3}xy^2+2xy-4y^2\)

Answer 2

a) \(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)

\(=\dfrac{1}{2}x^3-5x^2-x+10x+\dfrac{3}{2}x-15\)

\(=\dfrac{1}{2}x^3-5x^2+\dfrac{48}{5}x-15\)

b) \(\left(x^2y^2-\dfrac{1}{3}xy+2y\right)\left(x-2y\right)\)

\(=x^3y^2-2x^2y^3-\dfrac{1}{3}x^2y+\dfrac{2}{3}xy^2+2xy-4y^2\)