Question

\(2x^2-4x\)

\(3x^3+6x^2+3x\)

\(10x\left(x-y\right)-6x\left(y-x\right)\)

\(\left(x+1\right)^2-25\)

\(x^2+3x-y^2+3y\)

\(3x^2+5y-3xy-5x\)

\(x^2-7x-y^2+7y\)

\(3y^2-3z^2+3x^2\)

Answer 1

\(2x^2-4x=2x\left(x-2\right)\)

\(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)

\(10\left(x-y\right)-6x\left(y-x\right)=10\left(x-y\right)+6x\left(x-y\right)=\left(10+6x\right)\left(x-y\right)=2\left(x-y\right)\left(3x+5\right)\)\(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

\(x^2+3x-y^2+3y=\left(x-y\right)\left(x+y\right)+3\left(x+y\right)=\left(x+y\right)\left(x-y+3\right)\)

\(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)

\(x^2-7x-y^2+7y=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\)

\(3y^2-3z^2+3x^2=3\left(y^2-z^2+x^2\right)\)