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i) factorize 8p²+2p-8pq-2q ii) factorize 8x²+32xy+32y² iii) using results of i) and ii), factorize (8x²+32xy+32y²)+2x+4y-8(x+2y)(y-3)-2y+6?

2 個解答

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  • 3 年前
    最愛解答

    Sol

    i)

    8p^2+2p-8pq-2q

    =(8p^2-2p)-(8pq+2q)

    =2p(4p-1)-2q(4p+1)

    =(2p-2q)(4p+1)

    =2(p-q)(4p+1)

    ii)

    8x^2+32xy+32y^2

    =8(x^2+4xy+4y^2)

    =8(x+2y)^2

    iii)

    Set p=x+2y,q=y-3

    p-q=(x+2y)-(y-3)=x+y+3

    (8x^2+32xy+32y^2)+2x+4y-8(x+2y)(y-3)-2y+6

    =8(x+2y)^2+2(x+2y)-8(x+2y)(y-3)-2(y-3)

    =8p^2+2p-8pq-2q

    =2(p-q)(4p+1)

    =2(x+y+3)(4x+8y+1)

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  • 3 年前

    1.(8p+2)(p-q)

    2.(4x+6y)(2x+6y)

    =2(2x+3y)(x+3y)

    3.2(x+2y)(x+3y)+2(x+2y)-8(x+2y)(y-3)-2(y-3)

    =2(x+2y)(x+3y+1)-2(y-3)(4x-8y+1)

    =4(x+2y)(y+3)(x+3y+1)(4x+8y-1)

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