How do you multiply # (x-1)(x+1)(x-3)(x-5)#?

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Answer 1 · 2015-06-27T23:02:33

Consider each power of #x# in descending order and total up the possible combinations of coefficients to find:

#(x-1)(x+1)(x-3)(x-5) = x^4-8x^3+14x^2+8x-15#

Given #f(x) = (x-1)(x+1)(x-3)(x-5)#, consider each power of #x# in descending order and total up the combinations of coefficients:

#x^4# : #1*1*1*1 = 1#

#x^3# : #(-1*1*1*1)+(1*1*1*1)+(1*1*-3*1)+(1*1*1*-5) = -1+1-3-5 = -8#

#x^2# : #(-1*1*1*1)+(-1*1*-3*1)+(-1*1*1*-5)+(1*1*-3*1)+(1*1*1*-5)+(1*1*-3*-5) = -1+3+5-3-5+15 = 14#

#x# : #(-1*1*-3*1)+(-1*1*1*-5)+(-1*1*-3*-5)+(1*1*-3*-5) = 3+5-15+15 = 8#

#1# : #-1*1*-3*-5 = -15#

So #f(x) = x^4-8x^3+14x^2+8x-15#

Check:

#f(2) = 2^4-8*2^3+14*2^2+8*2-15#

#=16 - 64 + 56 + 16 - 15#

#=9#

#(2-1)(2+1)(2-3)(2-5) = 1*3*-1*-3 = 9#