How do you use FOIL to multiply $(x-2) (x^2-5x+3)$ ?

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Answer 1 · 2015-06-17T11:19:32

FOIL helps with multiplying two binomials, but is not applicable to multiplying a binomial by a trinomial.

Instead use distributivity to find:

$(x-2)(x^2-5x+3) = x^3-7x^2+13x-6$

$(x-2)(x^2-5x+3)$

$=x(x^2-5x+3)-2(x^2-5x+3)$

$=x^3-5x^2+3x-2x^2+10x-6$

$=x^3-(5x^2+2x^2)+(3x+10x)-6$

$=x^2-(5+2)x^2+(3+10)x-6$

$=x^2-7x^2+13x-6$

Alternatively, do what I do, look at each of the powers of $x$ in descending order in turn and total up the coefficients:

$x^3$ : $1*1 = 1$
$x^2$ : $(1*-5)+(-2*1) = -5-2 = -7$
$x$ : $(1*3)+(-2*-5) = 3+10 = 13$
$1$ : $-2*3 = -6$

Putting these together:

$x^3-7x^2+13x-6$

How do you use FOIL to multiply (x-2) (x^2-5x+3)(x-2) (x^2-5x+3)?