Question

the fraction of 3 k squared plus k minus 2 over 4 k minus 2 divided by the fraction of k squared plus 3 k plus 2 all over 2 k squared plus 5 k minus 3.

3k2+k−2/ 4k−2÷k2+3k+2/ 2k2+5k−3

Answer 1

Ill give a metal for who ever will just solve this problem.

Answer 2

\( \dfrac{3k^2+k−2}{4k−2} ÷ \dfrac{k^2+3k+2}{2k^2+5k−3} \)

The way you divide fractions is by multiplying the first one by the reciprocal of the second one. The reciprocal is fancy math talk for flipping a fraction.

That means your problem becomes:

\( =\dfrac{3k^2+k−2}{4k−2} \times \dfrac{2k^2+5k−3} {k^2+3k+2} \)

Now to actually multiply the fractions, you multiply the numerators together, and you multiply the denominators together.

\( =\dfrac{(3k^2+k−2)(2k^2+5k−3 ) }{(4k−2)(k^2+3k+2 )} \)

Now you need to reduce the fraction. To do that you need to factor every one of the 4 polynomials.

\( =\dfrac{\color{red}{(3k^2+k−2)}\color{green}{(2k^2+5k−3) } }{\color{brown}{(4k−2)}\color{blue}{(k^2+3k+2) }} \)

I wrote each polynomial in a different color so you can see what its factorization is below.

\( =\dfrac{\color{red}{(3k -2)(k + 1)}(\color{green}{2k - 1)(k +3) }) }{\color{brown}{2(2k−1)}\color{blue}{(k + 2)(k + 1)}} \)

Now we divide the numerator and denominator by all the common factors. This is what people call canceling out terms.

\( =\dfrac{\color{red}{(3k -2)(\cancel{k + 1})}(\color{green}{\cancel{2k - 1})(k +3) }) }{\color{brown}{2(\cancel{2k−1})}\color{blue}{(k + 2)(\cancel{k + 1})}} \)

The final answer is:

\( =\dfrac{(3k -2)(k +3) }{2(k + 2) }\)

You can multiply out the numerator and denominator, but there is no need for it.

Answer 3

Thank you very much!! Next time ill tag you to ask for help

Answer 4

You're welcome.