To put $(x - 1)(x + 2)x^2(x - 3)^3$ in standard form, we need to expand each part and simplify.

The given expression is: $(x - 1)(x + 2)x^2(x - 3)^3$

Step 1: Expand $(x - 3)^3$

Using the binomial expansion for $(x - 3)^3$, we get: $(x - 3)^3 = x^3 - 9x^2 + 27x - 27$

Step 2: Expand $(x - 1)(x + 2)$

Now, expand $(x - 1)(x + 2)$: $(x - 1)(x + 2) = x^2 + x - 2$

Step 3: Multiply the expansions

Now we combine everything: $(x^2 + x - 2) \cdot x^2 \cdot (x^3 - 9x^2 + 27x - 27)$

Step 3a: Multiply $(x^2 + x - 2) \cdot x^2$

$(x^2 + x - 2) \cdot x^2 = x^4 + x^3 - 2x^2$

Step 3b: Multiply $x^4 + x^3 - 2x^2$ with $(x^3 - 9x^2 + 27x - 27)$

Expanding term by term:

1. $x^4 \cdot x^3 = x^7$
2. $x^4 \cdot (-9x^2) = -9x^6$
3. $x^4 \cdot 27x = 27x^5$
4. $x^4 \cdot (-27) = -27x^4$
5. $x^3 \cdot x^3 = x^6$
6. $x^3 \cdot (-9x^2) = -9x^5$
7. $x^3 \cdot 27x = 27x^4$
8. $x^3 \cdot (-27) = -27x^3$
9. $-2x^2 \cdot x^3 = -2x^5$
10. $-2x^2 \cdot (-9x^2) = 18x^4$
11. $-2x^2 \cdot 27x = -54x^3$
12. $-2x^2 \cdot (-27) = 54x^2$

Adding all terms, we get: $x^7 - 8x^6 + 16x^5 - 9x^4 - 81x^3 + 54x^2$

Thus, the expression in standard form is: $x^7 - 8x^6 + 16x^5 - 9x^4 - 81x^3 + 54x^2$

Would you like more detailed steps or have any questions?

1. How do binomial expansions work for higher powers?
2. What is the difference between standard and factored form?
3. Why is the order of multiplication important in expansion?
4. How does expanding polynomials change the degree?
5. What are applications of polynomials in standard form?

Tip: When expanding, start by simplifying the smaller expressions first, then tackle the larger combinations.