Question

Algebra 2 help please? Two multiple choice questions

Answer 1

HI ::DD

Answer 2

Hi (:

Answer 3

@Missy_shay

Answer 4

For #7, factor everything first

Answer 5

And cancel off any common factors, then you need to find the restriction (which will be whatever x makes the denominator equal to zero)

Answer 6

Need help factoring it?

Answer 7

@cowboy 98 @precal @helpme7 @cwrw238

Answer 8

Okay well 7 factors as this. Now cancel off any common factors on the top and bottom.\[\Large \frac{ (x+2)(x+1) }{ (x-3)(x+1) }*\frac{ (x+3)(x+1) }{ (x+2)}\]

Answer 9

@Savannah_Noelle

Answer 10

So it becomes x^2+2/(x^2-3) * (x+3)/(x+2)?

Answer 11

@agent0smith

Answer 12

No once you have it factored like this \[\Large \frac{ (x+2)(x+1) }{ (x-3)(x+1) }*\frac{ (x+3)(x+1) }{ (x+2)}\]all you need to do is cancel the common factors on the top and bottom. Notice there's a x+2 on the top left and the bottom right, so cancel those off. And the x+1 on the bottom left and the top right

Answer 13

Ohhhh okay so you end up with (x-3) and (x+3) right?

Answer 14

\[\Large \frac{ (x+3)(x+1) }{(x-3)}\]

Answer 15

You still have one of the (x+1) on the top, since there was two x+1 factors on top, one on the bottom

Answer 16

Now can you tell what value of x will make the denominator (the bottom) equal to zero?

Answer 17

Okay would (x-3) and (x+3) cancel out and it would be 1?

Answer 18

No, they're not exactly the same, you can only cancel EXACT factors. x+3 and x+3 yes, x-3 and x-3 yes, x+3 and x-3, no.

All you have to do now is see what value of x will make the bottom equal to zero.

Answer 19

\[\Large \frac{ (x+3)(x+1) }{(x-3)}\]the restriction comes from when the denominator equals zero, the denominator is x-3, so

\[\Large x - 3 \ne 0\]

Answer 20

Just solve it the same way you would
x - 3 = 0

what do you get for x?

Answer 21

so it's positive 3

Answer 22

yep so \[\large x \ne 3\]

Answer 23

Okay I'm a little confused. How did you factor number 7?

Answer 24

For \[\Large x^2+3x+2 \]you need two numbers that multiply to 2, and add up to 3... so 2 and 1, since 2+1=3 and 2*1 = 2

So you factor it using those \[\Large x^2+3x+2=(x+2)(x+1)\]

Answer 25

Do the same with \[\large x^2+4x+3\]you need two numbers that multiply to 3, and add up to 4.

Answer 26

I'm really confused. Can you show me step by step how you turned x^2+3x+2/x^2-2x-3*x^2+4x+3x/x+2 Into (x+2)(x+1)/(x-3)(x+1)*(x+3)(x+1)/(x+2)?

Answer 27

Do each one one at a time. We've done the first \[\Large x^2+3x+2=(x+2)(x+1)\]Try this one, \[\large x^2+4x+3\]by finding two numbers that multiply to 3 and add up to 4

Answer 28

Once you have those two numbers, factor it like this (replace the question marks with the two numbers you found)\[\large x^2+4x+3 = (x+ ?)(x+?)\]

Answer 29

Ohhh okay I understand, thank you! I'm going to try number 8

Answer 30

Okay, for 8, first factor out the 3 out of this \[\Large 3x^2 + 12x+9 = 3(x^2+4x+3)\]now factor what is inside the brackets, just like we did above

Answer 31

(x+4)(x-1)

Answer 32

but 4 times -1 is -4, not 3.

You need two numbers that multiply to 3, and add up to 4.

Answer 33

(x-4)(x+1)

Answer 34

They still don't multiply to 3: -4 times 1 is -4

List all the factors of 3 (numbers that multiply to equal 3:
3 and 1
-3 and -1

which will add up to 4?

Answer 35

3 and 1

Answer 36

Yes, so use those. That's usually the best way to do these - list all the factors of a number, THEN see which two will add up to the one you need

Answer 37

Oh okay

Answer 38

you still have to keep that 3 out front though\[\large 3x^2 + 12x+9 = 3(x^2+4x+3) = 3(x+3)(x+1) \]

Answer 39

Now for the \[\large x^2-4\]it's easier to use a difference of two squares, \[\large a^2 - b^2 = (a+b)(a-b)\]since 4 is 2^2:
\[\large x^2 - 2^2 = (x+2)(x-2)\]

Answer 40

Ohhh okay is 4x+4 already simplified?

Answer 41

No, they both have a common factor of 4, try taking it out.

For example:
3x+9 = 3(x+3)

Answer 42

(x+1)?

Answer 43

You're missing something. Look at the example above to see what you're missing.

Answer 44

4(x+1)

Answer 45

Good. Now we have \[\large \frac{ 3(x+3)(x+1) }{ (x+2)(x-2) }\div \frac{ 4(x+1) }{ (x+2) }\]

Answer 46

Okay so now I cancel out common factors right?

Answer 47

In this case since it's division with a fraction, not multiplication, you first have to flip the fraction on the right, to make it multiplication. So flip the right fraction upside down: \[\large \frac{ 3(x+3)(x+1) }{ (x+2)(x-2) } * \frac{(x+2) }{ 4(x+1) }\]now cancel things

Answer 48

3(x+3)/(x-2)/4

Answer 49

Like this, right? Very good :)\[\large \frac{ 3(x+3)}{ (x-2)*4 } \]

Answer 50

yes :) Okay now what?

Answer 51

What value of x makes the denominator equal to zero? The denominator is 4(x-2) so \[\large 4(x-2) \ne 0\]

Answer 52

2?

Answer 53

Excellent :) \[\large x \ne 2\]

Answer 54

Thank you so much <333 Literally love you yo.

Answer 55

haha you're welcome <3