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Tha area of a triangle is 5x^3+19x2+6x-18 with length x+3. Using synthetic division,what is the width of the rectangle

A)5x^2+4x-6

B)5x^2+34x+108+306/x+3

C)5x^3+4x^2-6x

D)5x^2+34+108+306/x-3

Respuesta :

we know that

area of rectangle = length*width

Let's assume

length of rectangle is L

width of rectangle is W

[tex]A=L*W[/tex]

we are given

[tex]A=5x^3+19x^2+6x-18[/tex]

[tex]L=x+3[/tex]

now, we can find width

[tex]W=\frac{A}{L}[/tex]

now, we can plug values

[tex]W=\frac{5x^3+19x^2+6x-18}{x+3}[/tex]

we can use synthetic divison method

we get

[tex]W=\frac{5x^3+19x^2+6x-18}{x+3}=5x^2+4x-6[/tex]

[tex]W=5x^2+4x-6[/tex]

so, option-A...........Answer

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