f(x)=2x^3-7x^2+4x+4, prove that (x-2) is a factor and factorise f(x) completely
if (X-2) is a factor, f(2)=0
22^3-72^2+4*2+4= 16-28+8+4=0
therefore (x-2) is a factor.
2x^3-7x^2+4x+4=(x-2)(2x^2-3x-2)=(x-2)(2x+1)(x-2)
BP
if (X-2) is a factor, f(2)=0
22^3-72^2+4*2+4= 16-28+8+4=0
therefore (x-2) is a factor.
2x^3-7x^2+4x+4=(x-2)(2x^2-3x-2)=(x-2)(2x+1)(x-2)
