How do you multiply $(x - 1) (x - 2) (x - 3)$ $(x-1)(x-2)(x-3)$ ?

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How do you multiply $(x - 1) (x - 2) (x - 3)$ $(x-1)(x-2)(x-3)$ ?

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Answer 1 · 2018-06-11T06:17:53

$(x - 1) (x - 2) (x - 3) = x^{3} - 6 x^{2} + 11 x - 6$ $(x-1)(x-2)(x-3) = x^3-6x^2+11x-6$

Explanation:

It is helpful to know that:

$(x - α) (x - β) (x - γ)$ $(x-alpha)(x-beta)(x-gamma)$

$= x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ$ $= x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma$

In this identity the expressions forming the coefficients of $x^{k}$ $x^k$ are (apart from the alternating signs), the elementary symmetric polynomials in $α, β, γ$ $alpha, beta, gamma$ , namely:

$α + β + γ$ $alpha+beta+gamma$
$α β + β γ + γ α$ $alphabeta+betagamma+gammaalpha$
$α β γ$ $alphabetagamma$

With $α = 1$ $alpha=1$ , $β = 2$ $beta=2$ and $γ = 3$ $gamma=3$ , we find:

$α + β + γ = 1 + 2 + 3 = 6$ $alpha+beta+gamma=1+2+3=6$

$α β + β γ + γ α = (1 \cdot 2) + (2 \cdot 3) + (3 \cdot 1) = 2 + 6 + 3 = 11$ $alphabeta+betagamma+gammaalpha=(1 * 2) + (2 * 3) + (3 * 1) = 2+6+3=11$

$α β γ = 1 \cdot 2 \cdot 3 = 6$ $alphabetagamma = 1 * 2 * 3 = 6$

So:

$(x - 1) (x - 2) (x - 3) = x^{3} - 6 x^{2} + 11 x - 6$ $(x-1)(x-2)(x-3) = x^3-6x^2+11x-6$

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How do you multiply $(x - 1) (x - 2) (x - 3)$ $(x-1)(x-2)(x-3)$ ?

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