How do you foil $2 (x + 2) (x + 5)$ $2 (x+2) (x+5)$ ?

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How do you foil $2 (x + 2) (x + 5)$ $2 (x+2) (x+5)$ ?

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Answer 1 · 2015-08-10T11:11:02

FOIL can only be applied to the 2 binomials; the result of that multiplication can then be multiplied by $2$ $2$ to get:
$XXXX$ $color(white)("XXXX")$ $2 x^{2} + 14 x + 20$ $2x^2+14x+20$

Explanation:

Evaluating the product of two binomials using FOIL:

Sum
$XXXX$ $color(white)("XXXX")$ product of First terms
$XXXX$ $color(white)("XXXX")$ product of Outside terms
$XXXX$ $color(white)("XXXX")$ product of Inside terms
$XXXX$ $color(white)("XXXX")$ product of Last terms

For the given example: $2 (x + 2) (x + 5)$ $2(x+2)(x+5)$ :

We can apply FOIL to $(x + 2) (x + 5)$ $(x+2)(x+5)$ as
$XXXX$ $color(white)("XXXX")$ First terms: $x, x$ $x, x$ $XXXX$ $color(white)("XXXX")$ Product $= x^{2}$ $= x^2$
$XXXX$ $color(white)("XXXX")$ Outside terms: $x, + 5$ $x, +5$ $XXXX$ $color(white)("XXXX")$ Product $= + 5 x$ $= +5x$
$XXXX$ $color(white)("XXXX")$ Inside terms: $+ 2, x$ $+2, x$ $XXXX$ $color(white)("XXXX")$ Product $= + 2 x$ $= +2x$
$XXXX$ $color(white)("XXXX")$ Last terms: $+ 2, + 5$ $+2, +5$ $XXXX$ $color(white)("XXXX")$ Product $= + 10$ $= +10$

Sum
$XXXX$ $color(white)("XXXX")$ $= x^{2} + 5 x + 2 x + 10$ $= x^2+5x+2x+10$

$XXXX$ $color(white)("XXXX")$ $= x^{2} + 7 x + 10$ $= x^2+7x+10$

So
$XXXX$ $color(white)("XXXX")$ $(x + 2) (x + 5) = x^{2} + 7 x + 10$ $(x+2)(x+5) = x^2+7x+10$

and
$XXXX$ $color(white)("XXXX")$ $2 (x + 2) (x + 7) = 2 x^{2} + 14 x + 20$ $2(x+2)(x+7) = 2x^2+14x+20$

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How do you foil $2 (x + 2) (x + 5)$ $2 (x+2) (x+5)$ ?

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Explanation:

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