數學題3條(因式分解)

Question

1.a的4次方-b的4次方+c的4次方-d的4次方-2(a的2次方*c的2次方-b的2次方*d的2次方)+4ac(b的2次方+d的2次方)-4bd(a的2次方+c的2次方)=?

2.1+2x+3x的2次+4x的3次+5x的6次+4x的7次+3x的8次+2x的9次+x的10次=?

3.(x+1)(x+3)(x+5)(x+7)+15=?

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要有步驟！

Answer 1

1.)=[a^4+2a^2*c^2+c^4-4bd(a^2+c^2)+4b^2d^2]-[b^4+2b^2*d^2+d^4-4ac(b^2d^2)+4a^2*c^2]
=[(a^2+c^2)^2-4bd(a^2+c^2)+(2bd)^2]-[(b^2+d^2)-4ac(b^2+d^2)+(2ac)^2]
=[(a^2+c^2-(2bd)]^2*[(b^2+d^2)-(2ac)^2]^2
=(a^2+c^2-2bd-b^2-d^2+2ac)(a^2+c^2-2bd+b^2-d^2-2ac)
=[(a-c)^2+(b-c)^2][(a+c)^2-(b+d)^2]
=[(a-c)^2+(b-d)^2](a+c-b-d)(a+c+b+d)

2.)=1+2x+(x^2)+(2x^2)+(4x^3)+(2x^4)+(3x^4)+(6x^5)+(3x^6)+(2x^6)+(4x^7)+(2x^8)+(x^8 )+(2x^9)+(x^10)
=(1+2x+x^2)+2x^2(1+2x+x^2)+3x^4(1+2x+x^2)+2x^6(1+2x+x^2)+x^8(1+2x+x^2)
=(1+2x+x^2)[1+(2x^2)+(3x^4)+(2x^6)+(x^8)]
=(1+x)^2[(1+x^4+x^8+2x^2+2x^4+2x^6)]
=(1+x)^2(1+x^2+x^4)^2
=(1+x)^2[(1+2x^2+x^4)-x^2]^2
=(1+x)^2[(1+x)^2-x^2]^2
=(1+x)^2[(1+x^2-x^2)(1+x^2+x^2)
=(1+x)^2(x^2+x+1)(x^2-x+1)

3.=(x^2+8x+12)(x^2+8x+10)
=(6+x)(2+x)(x^2+8x+10)

Answer 2

Is this really correct?

Answer 3

3.(x+1)(x+3)(x+5)(x+7)+15=(x+2)(x+6)(x^2+8*x+10)