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We know that the area of the rectangle is $10$ square feet. The dimensions of a rectangle are its length and width. Recall that the area of a rectangle is length times width. $A=ℓw⇒10=ℓw $
We know that the $length$ of the table is one foot shorter than twice its width. Let's call the $width$ of the table $x.$ The length can be written as $ℓ=2x−1.$ $10=ℓw⇒10=(2x−1)(x) $
We now have an equation with one variable. If we can solve for the width $x,$ then we can find the length. Let's start by putting our equation in standard form.
Now we can factor the right hand side of the equation so that we can apply the Zero-Product Property.
Now that the equation is in factored form, let's apply the Zero-Product Property.
We can see that $x=25 $ and $x=-2.$ However, recall that $x$ represents width. Since it doesn't make sense to have a negative value as a width, $x=25 $ is the only correct value.

$0=2x_{2}−x−10$

Factor

WriteSumWrite as a sum

$0=2x_{2}+4x−5x−10$

FactorOutFactor out $2x$

$0=2x(x+2)−5x−10$

SplitIntoFactorsSplit into factors

$0=2x(x+2)−5x−(2)(5)$

FactorOutFactor out $5$

$0=2x(x+2)−5(x+2)$

FactorOutFactor out $2x−5$

$0=(2x−5)(x+2)$

$0=(2x−5)(x+2)$

Solve for $x$

ZeroProdPropUse the Zero Product Property

$0=2x−50=x+2 (I)(II) $

$5=2x0=x+2 $

$25 =x0=x+2 $

$25 =x-2=x $

$(I), (II):$ Rearrange equation

$x=25 x=-2 $