標題:
10點------factorization`!
(a) -8+18p-9p2 (b) x(4x-37)-30 (c) 15(z+3)2+13(+3)-2 (d) 2x2+9xyz+4y2z2 ThZ<3
最佳解答:
(a) -8+18p-9p2 =-(9p2-18p+8) =-(3p-2)(3p-4) (b) x(4x-37)-30 =4x2-37x-30 =(4x+3)(x-10) (c) 15(z+3)2+13(z+3)-2 =[(z+3)+1][15(z+3)-2] =(z+4)(15z+43) (d) 2x2+9xyz+4y2z2 =(2x+yz)(x+4yz)
其他解答:
a) you may use quad formular to try it, but remember to check the answer and don't omit the - sign (or simply, cross method if you can see it) ans: -(3x-2)(3x-4) b) expand it and do the same as a) ans: (x-10)(4x+3) c) should be 15(z+3)2+13(z+3)-2 instead, right? put z+3=x, u will get 15x2+13x-2 and the ans should be (15x-2)(x+1) back substitute x=z+3 into the original equation ans: (15z+43)(z+4) d) it should not be so easy to look from the polynomial right here at first, you must do it by observation (cross method), you will get the ans ans: (2x+yz)(x+4yz) hope u understand how it is done!
10點------factorization`!
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發問:(a) -8+18p-9p2 (b) x(4x-37)-30 (c) 15(z+3)2+13(+3)-2 (d) 2x2+9xyz+4y2z2 ThZ<3
最佳解答:
(a) -8+18p-9p2 =-(9p2-18p+8) =-(3p-2)(3p-4) (b) x(4x-37)-30 =4x2-37x-30 =(4x+3)(x-10) (c) 15(z+3)2+13(z+3)-2 =[(z+3)+1][15(z+3)-2] =(z+4)(15z+43) (d) 2x2+9xyz+4y2z2 =(2x+yz)(x+4yz)
其他解答:
a) you may use quad formular to try it, but remember to check the answer and don't omit the - sign (or simply, cross method if you can see it) ans: -(3x-2)(3x-4) b) expand it and do the same as a) ans: (x-10)(4x+3) c) should be 15(z+3)2+13(z+3)-2 instead, right? put z+3=x, u will get 15x2+13x-2 and the ans should be (15x-2)(x+1) back substitute x=z+3 into the original equation ans: (15z+43)(z+4) d) it should not be so easy to look from the polynomial right here at first, you must do it by observation (cross method), you will get the ans ans: (2x+yz)(x+4yz) hope u understand how it is done!
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