Question

因式分解(x²+3x+2)(x²+7x+12)-120

Answer 1

(x^2+3x+2)(x^2+7x+12)-120
=(x+1)(x+2)(x+3)(x+4)-120
=(x+1)(x+4)(x+2)(x+3)-120
=(x^2+5x+4)(x^2+5x+6)-120
设y=x^2+5x+5
原式
=(y-1)(y+1)-120
=y^2-1-120
=y^2-121
=(y+11)(y-11)
=(x^2+5x+16)(x^2+5x-6)
=(x^2+5x+16)(x-1)(x+6)

Answer 2

(x²+3x+2)(x²+7x+12)-120
=（x+1)(x+2)(x+3)(x+4)-120
=(x^2+5x+6)(x^+5x+4)-120
=（x^2+5x+6)^2-2(x^2+5x+6)-120
=(x^2+5x+6+10)(x^2+5x+6-12)
=(x^2+5x+16)(x^2+5x-6)
=(x^2+5x+16)(x-1)(x+6)

Answer 3

p(x) = (x²首首绝逗+3x+2)(x²者宏数+7x+12)-120 = (x+1)(x+2)(x+3)(x+4) - 2*3*4*5
p(1) = 0
p(-6) = 0
p(x) = (x-1)(x+6)(x^2 + 5x +16)

Answer 4

(x²棚哗+3x+2)(x²+7x+12)-120
=(X+2)(X+1)(X+3)(X+4)-120
=(x²+5X+4)(X2+5X+6)-120
=(x²返和升漏老+5X)²+10(x²+5X)-96
=(x²+5X+16)(x²+5X-6)
=(x²+5X+16)(X+6)(X-1)

Answer 5

x^2+5x-11)(x^2+5x+11)