Question
Simplify:
$\left(x^3-2 x^2+3 x-4\right)(x-1)-(2 x-3)\left(x^2-x+1\right)$
$\left(x^3-2 x^2+3 x-4\right)(x-1)-(2 x-3)\left(x^2-x+1\right)$
✓
Answer
To simplify, we will proceed as follows: $\left(x^3-2 x^2+3 x-4\right)(x-1)-(2 x-3)\left(x^2-x+1\right)$
$=\left[\left(x^3-2 x^2+3 x-4\right)(x-1)\right]-[(2 x-3)(x 2-x+1)]$
$=\left[x\left(x^3-2 x^2+3 x-4\right)-1\left(x^3-2 x^2+3 x-4\right)\right]-\left[2 x\left(x^2-x+1\right)-3\left(x^2-x+1\right)\right] \text { (Distributive law) }$
$=x^4-2 x^3+3 x^2-4 x-x 3+2 x^2-3 x+4-\left[2 x^3-2 x^2+2 x-3 x^2+3 x-3\right]$
$=x^4-2 x^3+3 x^2-4 x-x 3+2 x^2-3 x+4-2 x^3+2 x^2-2 x+3 x^2-3 x+3$
$=x 4-2 x^3-2 x^3-x^3+3 x^2+2 x^2+2 x^2+3 x^2-4 x-3 x-2 x-3 x+4+3 \text { (Rearranging) }$
$=x^4-5 x^3+10 x^2-12 x+7 \text { (Combining like terms) }$
Thus, the answer is $x^4-5 x^3+10 x^2-12 x+7$.
$=\left[\left(x^3-2 x^2+3 x-4\right)(x-1)\right]-[(2 x-3)(x 2-x+1)]$
$=\left[x\left(x^3-2 x^2+3 x-4\right)-1\left(x^3-2 x^2+3 x-4\right)\right]-\left[2 x\left(x^2-x+1\right)-3\left(x^2-x+1\right)\right] \text { (Distributive law) }$
$=x^4-2 x^3+3 x^2-4 x-x 3+2 x^2-3 x+4-\left[2 x^3-2 x^2+2 x-3 x^2+3 x-3\right]$
$=x^4-2 x^3+3 x^2-4 x-x 3+2 x^2-3 x+4-2 x^3+2 x^2-2 x+3 x^2-3 x+3$
$=x 4-2 x^3-2 x^3-x^3+3 x^2+2 x^2+2 x^2+3 x^2-4 x-3 x-2 x-3 x+4+3 \text { (Rearranging) }$
$=x^4-5 x^3+10 x^2-12 x+7 \text { (Combining like terms) }$
Thus, the answer is $x^4-5 x^3+10 x^2-12 x+7$.
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