用提公因式法计算

Question

4q(1-p)^3+2(p-1）^2 (3a-4b)(7a-8b)+(11a+2b)(8b-7a)

Answer 1

4q(1-p)^3+2(p-1）^2
=4q(1-p)^3+2(1-p）^2
=2(1-p)^2[2q(1-p)+1]
=2(1-p)^2(2q-2p+1)

(3a-4b)(7a-8b)+(11a+2b)(8b-7a)
= (3a-4b)(7a-8b)-(11a+2b)(7a-8b)
=(7a-8b)[(3a-4b)-(11a+2b)]
=(7a-8b)(-8a-6b)
=-2(7a-8b)(4a+3b)追问

为什么2（p-1）^2可以直接变成2(1-p)^2

追答

因为：(p-1)^2=[-(1-p)]^2=(1-p)^2

Answer 2

4q(1-p)^3+2(p-1）^2
=4q(1-p)^3+2(1-p)^2
=2(1-p)^2[2q(1-p)+1]
=2(1-p)^2(2q-2pq+1)

(3a-4b)(7a-8b)+(11a+2b)(8b-7a)

=(3a-4b)(7a-8b)-(11a+2b)(7a-8b)
=(7a-8b)[(3a-4b)-(11a+2b)]
=(7a-8b)(3a-4b-11a-2b)
=(7a-8b)(-8a-6b)
=-2(7a-8b)(4a+3b)追问

为什么2（p-1）^2可以直接变成2(1-p)^2

追答

(p-1)^2=[-(1-p)]^2=(1-p)^2

就是平常我们说的两个互为相反数的平方相等

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