(a)Expand the following expression. (i) (x-4)(x-6)= (ii) (x-2)(x-8)= (b)Hence,factorize(x-2)(x-4)(x-6)(x-8)+16. 求解答，please?

(a)


(i) (x - 4)(x - 6)

= x² - 6x - 4x + 24

= x² - 10x + 24

∴ (x - 4)(x - 6) = x² - 10x + 24........(1)


(ii) (x - 2)(x - 8)

= x² - 8x - 2x + 16

= x² - 10x + 16

∴ (x - 2)(x - 8) = x² - 10x + 16.........(2)


(b) (x - 2)(x - 4)(x - 6)(x - 8) + 16

= [(x - 4)(x - 6)][(x - 2)(x - 8)] + 16

= [x² - 10x + 24][x² - 10x + 16] + 16.........[From (1) & (2)]

= [(x² - 10x) + 24][(x² - 10x) + 16] + 16

= (x² - 10x)² + 16(x² - 10x) + 24(x² - 10x) + 384 + 16

= (x² - 10x)² + 40(x² - 10x) + 400

= (x² - 10x)² + (2)(20)(x² - 10x) + (20)²

= [(x² - 10x) + 20]²

= (x² - 10x + 20)²

(a)
(i) (x - 4)(x - 6) = x² - 10x + 24
(ii) (x - 2)(x - 8) = x² - 10x + 16

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策略 for (b)：
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- (b)題 starts with "Hence ...." ,明顯地指出 (b) 的 solution 是要利用 (a) 做出的 solution.

- From (a),
(i) (x - 4)(x - 6) = x² - 10x + ....
(ii) (x - 2)(x - 8) = x² - 10x + ....,
obviously see that both (x - 4)(x - 6) & (x - 2)(x - 8) arrive at x² - 10x + .... when expanded.
This gives me 「啟示」 to start with！
So, I rearrange (i) & (ii) as below:-
(i) (x - 4)(x - 6) = (x² -10x) +24
(ii) (x - 2)(x - 8) = (x² -10x) +16

Sol.：
(b) (x - 2)(x - 4)(x - 6)(x - 8) + 16
=[(x - 4)(x - 6)] [(x - 2)(x - 8)] + 16
= [(x² -10x) +24 ] [(x² -10x) +16 ] + 16
= (x² -10x)² +40(x² -10x)+ (24)16 + 16
= (x² -10x)² +40(x² -10x)+ (25)16
= (x² -10x)² +40(x² -10x)+ 400
= [(x² -10x) + 20]²
=(x² -10x +20)²