Toán Lớp 8: phân tích đa thức thành nhân tử: a, x(x+4)(x+6)(x+10)(+128) b, (4x+1)(12x-1)(3x+2)(x+1)-4 c, (x+3)^4+(x+1)^4-16

Question

Toán Lớp 8:  phân tích đa thức thành nhân tử: a, x(x+4)(x+6)(x+10)(+128) b, (4x+1)(12x-1)(3x+2)(x+1)-4 c, (x+3)^4+(x+1)^4-16

minhtamnguyet88 · Answer

a)   x (x+4)(x+6)(x+10) + 128    = [ x (x+10)] [ (x+4)(x+6)] +128    = (x^2 + 10x)(x^2 + 4x+  6x + 24) + 128    = (x^2+  10x)(x^2 + 10x +24) + 128    = [ (x^2 + 10x + 12) - 12] [ (x^2 + 10x + 12) + 12] + 128    = (x^2 + 10x + 12)^2 - 12^2 + 128   =  (x^2 + 10x + 12)^2 - 16    = (x^2 + 10x +12)^2 - 4^2   =  (x^2 + 10x + 12 - 4)(x^2 + 10x + 12 + 4)    = (x^2+ 10x + 8)(x^2 + 10x + 16)    = (x^2+ 10x + 8) (x^2 + 2x + 8x + 16)    = (x^2 + 10x + 8) [ x (x+2) + 8 (x+2)]    = (x^2 + 10x+8)(x+2)(x+8)   b)   (4x+1)(12x-1)(3x+2)(x+1) - 4   =  [ (4x+1) (3x+2)] [ (12x - 1)(x+1)] - 4   = (12x^2 + 8x + 3x + 2)(12x^2 + 12x - x - 1) - 4   =  (12x^2 + 11x + 2)(12x^2 + 11x - 1) - 4   =  [ (12x^2 + 11x + 0,5) + 1,5] [ (12x^2 + 11x + 0,5) - 1,5] - 4   =  (12x^2 + 11x + 0,5)^2 - (1,5)^2 - 4   =  (12x^2 + 11x + 0,5)^2 - 6,25    = (12x^2+  11x + 0,5)^2 - (2,5)^2    = (12x^2 + 11x + 0,5 - 2,5)(12x^2+ 11x + 0,5 + 2,5)   = (12x^2 + 11x-2)(12x^2 + 11x +3)   c)   (x+3)^4 + (x+1)^4 - 16    = (x+3)^4 + [ (x+1)^2 ]^2 - 4^2   =  (x+3)^4 +  [ (x+1)^2 - 4] [ (x+1)^2 + 4]   =  (x+3)^4 + (x^2 + 2x + 1 - 4)(x^2 + 2x + 1 + 4)    = (x+3)^4 + (x^2 + 2x - 3)(x^2 + 2x +5)    = (x+3)^4 + (x^2 - x + 3x - 3)(x^2 +2x+5)    = (x+3)^4+ [ x (x-1) + 3(x-1)] (x^2+  2x+5)    = (x+3)^4 + (x+3)(x-1)(x^2 + 2x+5)   = (x+3) [ (x+3)^3 + (x-1)(x^2 + 2x + 5)]   =  (x+3) ( x^3 + 9x^2 + 27x + 27 + x^3 + x^2 + 3x - 5)   =  (x+3) (2x^3 + 10x^2 + 30x + 22)   = (x+3) [ (2x^3 + 8x^2 + 22x)+ (2x^2 + 8x + 22)]    = (x+3) [ 2x (x^2 + 4x + 11) + 2 (x^2 + 4x + 11) ]    = (x+3) [ (2x+2)(x^2 + 4x + 11)]    = (x+3) . 2 (x+1)( x^2 + 4x + 11)

hienthucthu97 · Answer

Giải đáp+Lời giải và giải thích chi tiết:       a, x(x+4)(x+6)(x+10)+128     =(x^2+10x)(x^2+10x+24)+128     =(x^2+10x+12)^2-12^2+128     =(x^2+10x+12)^2-16     =(x^2+10x+16)(x^2+10x+8)     =(x+2)(x+8)(x^2+10x+8)     b, (4x+1)(12x-1)(3x+2)(x+1)-4     =(12x^2+11x+2)(12x^2+11x-1)-4     =(12x^2+11x+2)^2-3(12x^2+11x+2)-4     =(12x^2+11x+2)^2-4(12x^2+11x+2)+(12x^2+11x+2)-4     =(12x^2+11x+2)(12x^2+11x+2-4)+(12x^2+11x+2)-4     =(12x^2+11x+2-4)(12x^2+11x+2+1)     =(12x^2+11x-2)(12x^2+11x+3)     c, (x+3)^4+(x+1)^4-16    Đặt x+3=a,-x-1=b=>a+b=2   =>(x+3)^4+(x+1)^4-16   =a^4+b^4-(a+b)^4   =a^4+b^4-(a^4+4a^3b+6a^2b^2+4ab^3+b^4)   =-(4a^3b+6a^2b^2+4ab^3)   =-2ab(2a^2+3ab+2b^2)   =-2(x+3)(-x-1)[2(x+3)^2+3(x+3)(-x-1)+2(-x-1)^2]   =(x+3)(x+1)[2(x^2+6x+9)-3(x^2+4x+3)+2x^2+4x+2]   =(x+3)(x+1)(x^2+4x+11)