## anonymous one year ago Express the rational expression in lowest terms: 3x+4/15x^2+26x+8.

1. anonymous

by "lowest terms" you want to factor this?

2. anonymous

I think so, heres the question again with the choices:

3. anonymous

Oh, alright. so you want to factor $$\sf 15x^2+26x+8$$

4. anonymous

$\frac{4 x^2}{15}+29 x+8$

5. anonymous

i'd say your best choice would be to leave it as it is.,

6. anonymous

I"m not sure, but I just wanna know how to solve this

7. Nnesha

do you know how to factor the quadratic equations ???

8. anonymous

9. Nnesha

alright there are like 3 ways to factor the quadratic equation but the easy one is AC method you have to find two numbers when you multiply them you should get product of AC and when you add or subtract them you should get the middle term B $\huge\rm Ax^2+Bx+C$ A=leading coefficient B=middle term C=constant term

10. Nnesha

|dw:1442180776707:dw| multiply A C

11. anonymous

But @Nnesha the factored terms will not cancel out with the numerator..

12. Nnesha

yes one of the factor will cancel out with the numerator.

13. Nnesha

A=15 B=26 C=8 multiply 15 times 8= 120 now find two number when you multiply them u should get 120 and when u add or subtract them you should get 26

14. Nnesha

what are the two numbers whose sum is 26 and product is 120 ???

15. anonymous

Is it 6 and 20?

16. Nnesha

yes right now we have to apply group method |dw:1442182184861:dw| make a group of two terms now find take out the common factor from first red box

17. Nnesha

15x^2+20x what is common in these two terms ?

18. anonymous

I"m not sure?

19. Nnesha

what are the factors of 15 and 20 ?

20. Nnesha

nvm 5 times 3 =15 5 times 4 = 20 so 5,3 are factors of 15 and 5,4 are factors of 20 and x^2 can be written as x times x so $\huge\rm (\color{ReD}{5 · 3 ·x ·x} +\color{blue}{ 5 · 4 ·x})$ now can you tell me what's common in red and blue terms ?

21. anonymous

15 x squared is equal from the red, and 20x is blue?

22. Nnesha

yes it's 15x^2 +20x but i factored 15 , and 20 so you can find the common factor nvm so $\huge\rm (\color{ReD}{5} · 3 ·x ·\color{ReD}{x} +\color{red}{ 5} · 4 ·\color{ReD}{x})$ now as you can see 5 and x are common in both terms right ? so take them out $\huge\rm \color{ReD}{5x}(3x+4)$ understood ?

23. Nnesha

15x^2 = first term 20x = 2nd term just in case if you're not familiar with it :=)

24. anonymous

$15x^2+26x+8 \implies x^2+26x+120$$=(x+6)(x+20)$$=\left(x+\frac{6}{15}\right)\left(x+\frac{20}{15}\right)$$=\left(x+\frac{\cancel{6}2}{\cancel{15}5}\right)\left(x+\frac{\cancel{20}4}{\cancel{15}3}\right)$$\sf \text{cross multiply} : (5x+2)(3x+4)$$=\frac{3x+4}{(5x+2)(3x+4)}$ So what would eliminating $$\sf 3x+4$$ from the top and bottom give you?

25. Nnesha

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