Answer:
The width of the area model is equal to
[tex](2x^2-x+3)\ units[/tex]
Step-by-step explanation:
The complete question is
Todor was trying to factor 10x^2-5x+15 he found the greatest common factor of these terms was 5 what is the width
we know that
The area of a rectangular model is given by the formula
[tex]A=LW[/tex] ----> equation A
where
L is the length
W is the width
we have
[tex]A=10x^2-5x+15[/tex]
Factor the expression
[tex]A=5(2x^2-x+3)[/tex]
substitute the value of the Area in the equation A
[tex]5(2x^2-x+3)=LW[/tex]
In this problem
The greatest common factor of these terms is the length (L=5 units)
so
we can say that the width is equal to (2x^2-x+3)
therefore
The width of the area model is equal to
[tex](2x^2-x+3)\ units[/tex]
