When we are given the area of a rectangle, we generally need to use the formula for the area of a rectangle. In this case we need to write expressions for the length, $ℓ,$ and width, $w .$

Verbal Expression	Algebraic Expression
the length is $10 ft$ more than the width	$ℓ = 10 + w$

Now that we have an algebraic expression, we can substitute. Then, let's see where we can go from there.

A = ℓ \cdot w

SubstituteII

$A = 144$ , $ℓ = 10 + w$

144 = (10 + w) w

▼

Simplify

Distr

Distribute $w$

144 = 10 w + w^{2}

SubEqn

$LHS - 144 = RHS - 144$

0 = 10 w + w^{2} - 144

CommutativePropAdd

Commutative Property of Addition

0 = w^{2} + 10 w - 144

RearrangeEqn

Rearrange equation

w^{2} + 10 w - 144 = 0

Now, we have a quadratic equation that is equal to zero. Let's look for two factors of

- 144

that have a sum of

10 .

Since

a c

is negative, the sign of the factors will be different. Furthermore, since

b

is positive, the factor with the larger magnitude will be positive too.

Factors	Sum of factors
$- 1, 144$	$143$
$- 2, 72$	$70$
$- 3, 48$	$45$
$- 4, 36$	$32$
$- 6, 24$	$18$
$- 8, 18$	$10$

We have found our factors that have a sum of

10 .

Since our leading coefficient is

1,

we can use

- 8

and

18

directly as part of the factoring.

w^{2} + 10 w - 144 = (w - 8) (w + 18)

Let's set each factor equal to zero and solve for

w .

(w - 8) (w + 18) = 0

▼

Solve for

w

ZeroProdProp

Use the Zero Product Property

w - 8 = 0 w + 18 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 8 = RHS + 8$

w = 8 w + 18 = 0

SubEqn

$(II):$ $LHS - 18 = RHS - 18$

w = 8 w = - 18

Since

w

represents a measurement of distance, we can ignore the negative solution and say the width of the rectangle is

8 ft .

To find the length, we can add

10

to the width and get

18 ft .

Exercise