Math question

Well, you know that the area of a rectangle is L x W, right? So what times the length (X+4) is (X^2+6X+8)? You can factor X^2+6X+8 into (X+4)(X+2), so the width must be ...

I need to solve it very sure,,,

What are you unsure about? What's your guess?

At last some real "algebra" which you are now calling "math" !

How do you get the area of a rectangle? You multiply the length by the width right? So if you know the area and the length, to get the width you divide the area by the length, right?

Now here the area is given by x^2+6x+8 in^2 and the length is x+4 in.

Now the area is given as a quadratic and the length is a linear (x+4)---so the width is another linear expression of x plus some constant, right? I'll call this constant D. OK!

So take the length (x+4) and multiply it times the width (x+D)

(x+4 )(x+D)=x^2+4x+Dx+4D=x^2+(4+D)x+4D

Now compare this with the known area (x^2+6x+8) and see if D can be determined, right?

so 4+D=6 and 8=4D

lets start with 8 .
Well 8=2*2*2=2*4
So D could be 2, right?

Check D with the other
4+D=4+2=6 OK now D=2 works, right?

and x^2+6x+8=(x+4)(x+2)

If the length is (x+4) in, then the width is (x+2) in, right?

Rap