可以幫我因式分解這兩題嗎? (1)(x^2-y^2)^2-8(x^2+y^2)+16 (2)(2x+1)(x^2-x+3)+2x(x-6) ps:我要過程?

(1)
(x²-y²)²-8(x²+y²)+16
= [(x²-y²)²+16] -8(x²+y²)
= [(x²-y²) - 4]² + 8(x²-y²) -8(x²+y²) . . . . . . . . . . . .(完全平方)
= (x²-y² - 4 )² -16y²
= (x²-y² - 4 )² - (4y)²
= [(x²-y² - 4 ) + 4y] [(x²-y² - 4 ) - 4y] . . . . . . . . . . . .{∵ A² - B² = (A+B)(A-B) }
= (x² - y² + 4y - 4) (x²- y² - 4y - 4)
= [x² - (y²- 4y + 4)] [(x²- (y² + 4y + 4)]
= [x² - (y- 2)²] [(x²- (y+ 2)²]
= [x + (y- 2)] [x - (y- 2)] [x + (y+2)] [x - (y+2)]
= (x + y - 2) (x - y + 2) (x + y + 2) (x - y - 2)

(2) 設 f(x) = (2x+1)(x²-x+3)+2x(x-6)

(2x+1)(x²-x+3)+2x(x-6)
.^^^ . . . . . . . . . ^^^
= (2x) [(x²-x+3)+(x-6)] + (x²-x+3)
. .^^^
= (2x)(x²-3) + (x²-x+3)
= 2x³ + x² -7x +3

∵ f(1/2) = 2(1/2)³ +(1/2)² -7(1/2) +3 = 1/4 +1/4 -7/2 +3 = 0
∴ (x-1/2) 是 f(x) 的因數

設 f(x) = (x-1/2) q(X)
=> 2x³ + x² -7x +3 = (x-1/2) q(X)
=> 4x³ + 2x² -14x +6 = (2x-1) q(X)
=> q(x) = (4x³ + 2x² -14x +6) ÷ (2x-1) = 2x² +2x - 6 - - - - - - - - - - - (長除法)

∴ f(x) = 2x³ + x² -7x +3 = (x-1/2)(2x² +2x - 6) = (2x-1)(x² +x - 3)