Question

Can someone please help me with this math question?
Please? I will fan and give a medal!!

Answer 1

Please help me to simplify-
x^2 - 3x - 4
--------------
x - 4

Answer 2

factor the numerator, then cancel the common factor of \(x-4\)

Answer 3

to factor the numerator, find two numbers that...

a) multiply to -4 (last term)
AND
b) add to -3 (middle coefficient)

These two numbers will determine how the factorization goes

Answer 4

For example

x^2 + 5x + 6 factors to (x+3)(x+2)

since
3*2 = 6
3+2 = 5

Answer 5

@jim_thompson5910
What do you mean find two numbers that...
a) multiply to -4 (last term)
AND
b) add to -3 (middle coefficient)

Answer 6

Do you see how x^2+5x+6 factors to (x+3)(x+2) ?

Answer 7

Yes sir

Answer 8

notice how 3*2 = 6 and how 3+2 = 6

Answer 9

so how would you factor something like x^2+10x+24 ?

Answer 10

Yes sir, I'm trying to figure it out now what two numbers multiple to -4

Answer 11

-2*2 = -4

Answer 12

-2*2 = -4 is true

but -2 plus 2 = 0 when we want the two numbers to add to -3

Answer 13

Yes sir I'm trying to figure that out, but I'm not very good with negatives

Answer 14

what's another way to multiply to -4?

Answer 15

-4*1=-4
-4+1=-3
I didn't even think of this! The most simplest way!

Answer 16

very good

Answer 17

the two numbers are -4 and +1 so x^2-3x-4 factors to (x-4)(x+1)

Answer 18

Ok then what?

Answer 19

So we factor anything we can. Then we cancel out the common factors. In this case, a pair of (x-4) factors cancel out.

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

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

\[\Large \frac{x^2 - 3x - 4}{x-4} = x+1\]

where \[\Large x \ne 4\]

Answer 20

So,
\[\Large \frac{x^2 - 3x - 4}{x-4}\]
simplifies to
\[\Large x+1\]
where \[\Large x \ne 4\]

Answer 21

Thank you so much!!! I appreciate your help!!