First part :

x³ + 3x - x² - 3

= x(x² + 3) - (x² + 3)

= (x - 1)(x² + 3)

Thank you J! Would you be able to help on the 2nd part which is more tricky? Again thank you!

Second part :

(x² - 3)³ - (2 - x²)² + 3(x² - 3)

= (x² - 3)³ - (x² - 2)² + 3(x² - 3)

= (x² - 3)³ - ( (x² - 3) + 1)² + 3(x² - 3)

= (x² - 3)³ - ( (x² - 3)² + 2(x² - 3)(1) + 1² ) + 3(x² - 3)

= (x² - 3)³ - ( (x² - 3)² + 2x² - 6 + 1) + 3(x² - 3)

= (x² - 3)³ - ( (x² - 3)² + 2x² - 5) + 3(x² - 3)

= (x² - 3)³ - ((x² - 3)² + 3 + 2x² - 8) + 3(x² - 3)

= (x² - 3)³ - (x² - 3)² - 3 - 2x² + 8 + 3(x² - 3)

= (x² - 3)³ + 3(x² - 3) - (x² - 3)² - 3 - 2x² + 8

= ((x² - 3) - 1)((x² - 3)² + 3) - 2(x² - 4)

(Using the first part's result. Here we substitute x² - 3 for x)

= (x² - 4)(x⁴ - 6x² + 9 + 3) - 2(x² - 4)

= (x² - 4)(x⁴ - 6x² + 9 + 3 - 2)

= (x² - 4)(x⁴ - 6x² + 10)

= (x⁴ - 6x² + 10)(x² - 4)

= (x⁴ + (-6)x² + 10)(x² + (-4))

Therefore , A = -6, B = 10, C = -4

Edited. You may skip some steps if necessary

Wow thank you J!!

Welcome. The trick is to notice that the terms with power 3 and power 1 are already having the correct sign and coefficient.


So all we need to do is to manipulate the -(2 - x²)² and then create a (x² - 3)² from it by rewriting some of its terms as sum/difference of more terms.

First part also tests you to realise that (2 - x²)² can be rewritten as (x² - 2)²

Since (a - b)²
= a² - 2ab + b²
= b² - 2ab + a²
= (b - a)²

Or

(a - b)²
= ((-1)(b-a))²
= (-1)²(b-a)²
= (b - a)²


Likewise, (a+b)² = (-a-b)² = (b+a)² = (-b-a)²