## KE8717504 3 years ago consider the rectangle below. It's area is 5a^2+21ab+4b^2 and it's length is 2a+8b. find it's width w, in terms of a and b, by dividing it's area by it's length.

1. KE8717504

|dw:1347684283899:dw|

2. KE8717504

@jim_thompson5910

3. jim_thompson5910

what do you have so far

4. KE8717504

I'm just not sure how to work this.

5. KE8717504

or how to go about starting it?

6. jim_thompson5910

Use the idea that Area = Length * Width or A = LW Solve for W W = A/L

7. jim_thompson5910

Then make the proper substitutions. Then divide and simplify.

8. KE8717504

@jim_thompson5910 this is what i get when i do that. 2ab^2+5a^2+21ab+8b

9. KE8717504

That is diving the Area by the length

10. jim_thompson5910

simplify (5a^2+21ab+4b^2)/(2a+8b)

11. KE8717504

5a+b/2 ?

12. KE8717504

would that make my width?

13. jim_thompson5910

Can you factor 5a^2+21ab+4b^2 at all?

14. KE8717504

yes into (a+4b)(5a+1b) right?

15. KE8717504

but u still get the same answer right 5a+b/2 right?

16. KE8717504

@jim_thompson5910

17. Algebraic!

It's right.

18. jim_thompson5910

Did you mean $\Large \frac{5a+b}{2}$ ?

19. KE8717504

@jim_thompson5910 yea

20. jim_thompson5910

then yes, you are correct

21. KE8717504

I should get a learners metal on that one :0

22. KE8717504

Thanks !!