Question

The area of a rectangle is 5x^3+19^2+6-18 with length x + 3. Using synthetic division, what is the width of the rectangle?

Answer 1

Answer:

A. "5x^2 + 4x - 6"

Step-by-step explanation:

5x^2 + 4x - 6

Answer 2

Answer:

The width of the given rectangle is $5x^2+4x-6$ units

Therefore $w=5x^2+4x-6$ units

Step-by-step explanation:

GIven that the area of the rectangle is $5x^3+19x^2+6x-18$ and length is (x+3)

To find the width of the given rectangle

Area of the rectangle $A=lw$

$w=\frac{A}{l}$

$w=\frac{5x^3+19x^2+6x-18}{x+3}$

Solving the equation by synthetic division

-3_|  5     19      6     -18

       0     -15    -12     18

________________________

       5       4      -6      0

Therefore 5x^2+4x-6=0

Therefore $w=5x^2+4x-6$ units

Therefore the width of the given rectangle is $5x^2+4x-6$ units