Question

2x^(2)-7x-4/x^(2) * 8x^(2)-28x+12/4x^(2)-1

Multiply

Answer 1

@Hero

Answer 2

Use the drawing pad please.

Answer 3

ok

Answer 4

\[\frac{ 2x^{2}-7x-4 }{ x ^{2} }\times \frac{ 8x ^{2}-28x+12 }{ 4x ^{2}-1 }\]

Answer 5

@Mertsj

Answer 6

Did you try factoring the numerators yet?

Answer 7

let me do it reak quick n check

Answer 8

Now cancel factors of one, then simplify

Answer 9

all my sign were opposite than yours

Answer 10

how do i simplify

Answer 11

To simplify:
1. Cancel all factors of 1
2. Combine the product of both fractions into one fraction

Answer 12

\[\frac{ x+4 }{ x ^{2} } \times \frac{ 4(x-3) }{ 2x+1 }\] this?

Answer 13

You can only eliminate one set of factors of one at a time

Answer 14

You can't eliminate every (2x - 1) you see. That's not how it works.

Answer 15

Also, you need to combine everything into just one fraction.

For example, if you have

\[\frac{a}{b}×\frac{c}{d}\]


When you combine that into one fraction, you'll have

\[\frac{ac}{bd}\]

Answer 16

When you simplify the product of two fractions, you want to express your answer in the form of one fraction with no common factors.

A factor of one is a fraction such that you have the same number in the numerator and denominator:

\[\frac{a}{a}\]

You can only eliminate one common factor at a time.

If you have

\[\frac{aa}{a}\] Then you simplify as follows:

\[\frac{aa}{a} = \frac{a\cancel{a}}{\cancel{a}} = a\]

Answer 17

Notice that you don't cancel all the a's. Only the factors of one

Answer 18

Also

\[\frac{aa}{a} = a \times \frac{a}{a} = a \times 1 = a\]

Answer 19

so whats the answer

Answer 20

@Hero

Answer 21

...

Answer 22

Tell me what you think the answer is first.

Answer 23

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

Answer 24

@Hero

Answer 25

Okay, I made a sign error earlier. I apologize.

Answer 26

After factoring you should get:

\[\frac{(2x+1)(x-4)}{x^2} \dot\ \frac{4(2x-1)(x-3)}{(2x+1)(2x-1)}\]

Answer 27

so my answer wasnt right

Answer 28

Next rewrite in a way so that the factors of one are lined up one on top of the other:

\[\frac{(2x+1)(2x-1)}{(2x+1)(2x-1)} \dot\ \frac{4(x-4)(x-3)}{ x^2}\]

Then cancel factors of 1:

\[\frac{\cancel{(2x+1)(2x-1)}}{\cancel{(2x+1)(2x-1)}} \dot\ \frac{4(x-4)(x-3)}{ x^2}\]

Now you're left with the correct result:

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

Answer 29

that it?

Answer 30

Yes, that should be it. But sometimes, they want the expanded form.

Answer 31

You'll have to multiply the numerator in order to get that.

Answer 32

thats not an option on a b c or d

Answer 33

I just explained to you that you might have to multiply the numerator in order to get the expanded form.

Answer 34

Try it and see

Answer 35

idk how

Answer 36

A) \[\frac{ x-16 }{ x+3 }\] B)\[\frac{ x-4 }{ x+3 }\] c)\[\frac{ 4(x-4) }{ x+3 }\] D)\[\frac{ 4x-4 }{ x+3 }\]

Answer 37

You must have posted the original problem incorrectly. There's no way you get x + 3 in the denominator.

Answer 38

Double check that that original question and the answer choices are for the same problem.