Question

Can someone please help

Simplify quantity 4 x squared plus 12 x minus 16 all over quantity 2x plus 10 over quantity 6 x plus 24 over quantity x squared plus 9x plus 20

Answer 1

\[\Large \frac{\frac{4x^2 + 12x - 16}{2x + 10}}{\frac{6x + 24}{x^2 + 9x + 20}}\]

like that?

Answer 2

yes

Answer 3

First thing is first...when a fraction is being divided by another fraction...we can take the bottom fraction, flip it, and turn this to multiplication

\[\large \frac{4x^2 + 12x - 16}{2x + 10} \times \frac{x^2 + 9x + 20}{6x + 24}\]

Answer 4

ok got it then what ?

Answer 5

Now we begin to simplify...

lets focus on the first fraction we have

\[\large \frac{4x^2 + 12x - 16}{2x + 10}\]

What can be factored out of that numerator? (what number can each term be divided by?)

Answer 6

2

Answer 7

While 2 IS correct, it also looks like 4 is correct too right?

\[\large \frac{4(x^2 + 3x - 4)}{2x + 10}\] right?

now what number can be divided out of the denominator?

Answer 8

2 ?

Answer 9

Right...

so we will have

\[\large \frac{4(x^2 + 3x - 4)}{2(x + 5)}\]

got that?

Answer 10

ok yes

Answer 11

Alright, so 1 more thing about this fraction

Look what we have

\[\large \frac {\color\red{4}(x^2 + 3x - 4)}{\color \red{2}(x + 5)}\]

what is 4 divided by 2?

Answer 12

2

Answer 13

\[\large \frac{2(x^2 + 3x - 4)}{(x + 5)}\]

all that make sense?

Answer 14

yes it does

Answer 15

Good, now onto the second fraction

\[\large \frac{x^2 + 9x + 20}{6x + 24}\]

Lets begin...start with the denominator, what number can be divided out?

Answer 16

6

Answer 17

Good

\[\large \frac{x^2 + 9x + 20}{6(x + 4)}\]

Now how do we simplify the numerator?

Answer 18

We can factor it,

Ask yourself

What numbers multiply to make 20 AND ALSO add to make 9??

Answer 19

4 and 5

Answer 20

Correct...so that will breakdown into

\[\large \frac{(x + 4)(x + 5)}{6(x + 4)}\]

make sense?

Answer 21

yes .. so what would be the answer ?

Answer 22

Well we can still simplify the second fraction...

look what we have there

\[\large \frac{\color \red{(x + 4)}(x + 5)}{6\color \red{(x + 4)}}\]

They are on both the top AND bottom...so

\[\large \frac{\cancel{\color \red{(x + 4)}}(x + 5)}{6\cancel{\color \red{(x + 4)}}}\]

So we have a simplifed 2nd fraction of

\[\large \frac{(x + 5)}{6}\]

did all that make sense? Almost done

Answer 23

yes

Answer 24

Alright, we simplified the first fraction, and just now the second, now we just neeed to multiply them

\[\large \frac{2(x^2 + 3x - 4)}{(x + 5)} \times \frac{(x + 5)}{6}\]

First thing is first, combine them over 1 fraction

\[\large \frac{2(x^2 + 3x- 4)(x + 5)}{6(x + 5)}\]

make sense?

Answer 25

yes

Answer 26

alright, well we have something on top AND bottom, so those cancel

\[\large \frac{2(x^2 + 3x - 4)\cancel{(x + 5)}}{6\cancel{(x + 5)}}\]

Leaving us with

\[\large \frac{\color \red{2}(x^2 + 3x - 4)}{\color \red{6}}\]

the numbers bolded can be divided too, what is 2/6?

Answer 27

.33

Answer 28

Or \(\large \frac{1}{3}\) right?

Answer 29

yes

Answer 30

Alright so what we will have is

\[\large \frac{(x^2 + 3x- 4)}{3}\]

and that is our simplified problem! (technically it CAN be simplified 1 more step but it doesnt really change anything)

Answer 31

ok thank you so much

Answer 32

No problem!