Question

30题B和31题B怎么算（要步骤）

Answer 1

30.
(a)
x²-4x-21=(x+3)(x-7)
4x²+11x-3=(x+3)(4x-1)

(b)以(a)为基础：
原式
=2x(x²-4x-21)+4(4x²+11x-3)
=2x(x+3)(x-7)+4(x+3)(4x-1)
=2(x+3)[x(x-7)+2(4x-1)]
=2(x+3)(x²+x+2)
=2(x+3)(x+2)(x-1)

31.
(a)
立方差公式：a³-b³=(a-b)(a²+ab+b²)
8p³-27q³=(2p)³-(3q)³=(2p-3q)(4p²+6pq+9q²)

(b)以(a)为基础：
原式
=(8p³-27q³)+(12p²q-18pq²)
=(2p-3q)(4p²+6pq+9q²)+6pq(2p-3q)
=(2p-3q)[(4p²+6pq+9q²)+6pq]
=(2p-3q)(4p²+12pq+9q²)
=(2p-3q)(2p+3q)²

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Answer 2

30.b
原式
=2(x²-4x-21)+4(4x²+11x-3)
=2(x+3)(x-7)+4(x+3)(4x-1)
=2(x+3)[(x-7)+2(4x-1)]
=2(x+3)(x-7+8x-2)
=2(x+3)(9x-9)
=18(x+3)(x-1)

31.b
原式
=(8p³-27q³)+(12p²q-18pq²)
=(2p-3q)(4p²+6pq+9q²)+6pq(2p-3q)
=(2p-3q)[(4p²+6pq+9q²)+6pq]
=(2p-3q)(4p²+12pq+9q²)
=(2p-3q)(2p+3q)²

Answer 3

30 原式=2x(x^2-4x-21)+4(4x^2+11x-3)=2x(x-7)(x+3)+4(x+3)(4x-1)=2(x+3)(x+2)(x-1)
31 原式=8p^3-27q^3+12p^2q-18pq^2=(2p-3q)(4p^2+6pq+9q^2)+6pq(2p-3q)=(2p-3q)(2p+3q)^2