Question

1)(2x-1)(4x+2x+1)-4x(2x²-3)при x=0.5
2)x(x+2)(x-2)-(x-3)(x²+3x+9)при х=1/4​

Answer 1

$\displaystyle \tt 1) \: (2x-1)(4x+2x+1)-4x(2x^2-3)=(2x-1)(6x+1)-8x^3+12x=12x^2+2x-6x-1-8x^3+12x=12x^2+8x-1-8x^3=\bold{-8x^3+12x^2+8x-1}$

$\displaystyle \tt -8\cdot0,5^3+12\cdot0,5^2+8\cdot0,5-1=-1+3+4-1=5$

$\displaystyle \tt 2) \: x(x+2)(x-2)-(x-3)(x^2+3x+9)=x(x^2-4)-(x^3-27)=x^3-4x-x^3+27=\bold{-4x+27}$

$\displaystyle \tt -4\cdot\frac{1}{4}+27=-1+27=26$