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標題:
利用平方差的恆等式,因式分解下列各多項式。
發問:
利用平方差的恆等式,因式分解下列各多項式。 1.36a^2-25b^2 2.(a+b)^2-c^2 3.3x^3-27x 4.4x(y+z)^2-36x 5.p^2-q^2-4pr-4qr 6.4m^2-n^2+6m+3n
1.36a^2-25b^2 =(6a+5b)(6a-5b) 2.(a+b)^2-c^2 =(a+b+c)(a+b-c) 3.3x^3-27x =3x(x^2-9) =3x(x+3)(x-3) 4.4x(y+z)^2-36x =4x[(y+z)^2-9] =(y+z+3)(y+z-3) 5.p^2-q^2-4pr-4qr =(p+q)(p-q)-4(p+q) =(p+q)(p-q-4) 6.4m^2-n^2+6m+3n =(2m+n)(2m-n)+3(2m+n) =(2m+n)(2m-n+3) 2010-02-19 21:58:48 補充: 4.4x(y+z)^2-36x =4x[(y+z)^2-9] =4x(y+z+3)(y+z-3) 2010-02-19 22:14:09 補充: Thank you tonyleung052 for reminding me. 5.p^2-q^2-4pr-4qr =(p+q)(p-q)-4r(p+q) =(p+q)(p-q-4r)
其他解答:
36a^2-25b^2 (6a-5b)(6a+5b) (a+b)^2-c^2 (a+b-c)(a+b+c) 3x^3-27x 3x(x^2-9) 3x(x+3)(x-3) 4x(y+z)^2-36x 4x((y+z)^2-9) 4x(y+z-3)(y+z+3) p^2-q^2-4pr-4qr (p^2-q^2)-4r(p+q) (p+q)(p-q)-4r(p+q) (p+q)(p-q-4r) 4m^2-n^2+6m+3n (4m^2-n^2)+3(2m+n) (2m+n)(2m-n)+3(2m+n) (2m+n)(2m-n+3)|||||001wingkei : You made mistake in Q5
利用平方差的恆等式,因式分解下列各多項式。
發問:
利用平方差的恆等式,因式分解下列各多項式。 1.36a^2-25b^2 2.(a+b)^2-c^2 3.3x^3-27x 4.4x(y+z)^2-36x 5.p^2-q^2-4pr-4qr 6.4m^2-n^2+6m+3n
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最佳解答:1.36a^2-25b^2 =(6a+5b)(6a-5b) 2.(a+b)^2-c^2 =(a+b+c)(a+b-c) 3.3x^3-27x =3x(x^2-9) =3x(x+3)(x-3) 4.4x(y+z)^2-36x =4x[(y+z)^2-9] =(y+z+3)(y+z-3) 5.p^2-q^2-4pr-4qr =(p+q)(p-q)-4(p+q) =(p+q)(p-q-4) 6.4m^2-n^2+6m+3n =(2m+n)(2m-n)+3(2m+n) =(2m+n)(2m-n+3) 2010-02-19 21:58:48 補充: 4.4x(y+z)^2-36x =4x[(y+z)^2-9] =4x(y+z+3)(y+z-3) 2010-02-19 22:14:09 補充: Thank you tonyleung052 for reminding me. 5.p^2-q^2-4pr-4qr =(p+q)(p-q)-4r(p+q) =(p+q)(p-q-4r)
其他解答:
36a^2-25b^2 (6a-5b)(6a+5b) (a+b)^2-c^2 (a+b-c)(a+b+c) 3x^3-27x 3x(x^2-9) 3x(x+3)(x-3) 4x(y+z)^2-36x 4x((y+z)^2-9) 4x(y+z-3)(y+z+3) p^2-q^2-4pr-4qr (p^2-q^2)-4r(p+q) (p+q)(p-q)-4r(p+q) (p+q)(p-q-4r) 4m^2-n^2+6m+3n (4m^2-n^2)+3(2m+n) (2m+n)(2m-n)+3(2m+n) (2m+n)(2m-n+3)|||||001wingkei : You made mistake in Q5
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