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:

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

$=16 - 64 + 56 + 16 - 15$

$=9$

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

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