Question

Okay, so if i have x^2+4x-72=5x, i can subtract the 5x to get x^2-1x-72=0, right? How do you find the other numbers to go in the ( ) with the x?

Answer 1

Here's what I do. Look at the third number -72

Answer 2

Okay...

Answer 3

now the hard part, find its factors
1 x 72
2 x 36

Answer 4

3*26
4*18

Answer 5

6*12
8*9

Answer 6

Now because we have negative 72,(-72) find a pair when you subtract them, give you the second number in your equation ( -1 in front of the x)

Answer 7

as before, the two numbers are 8 and -9. So (x+8)(x-9), you get x=-8, x=9.

Answer 8

Oh! So you just find what factors of the last number give you the second number, then you switch the signs?

Answer 9

As you notice 8-9= -1 so those are our numbers
so the factors will be (x+8)(x-9)

Answer 10

and if you are not sure. double check the answer by multiply it out
(x+8)(x-9) = x^2 +8x -9x -72 = x^2 -x -72

Answer 11

Now if the 3rd number were positive, we would have to find the sum of a pair
for example,
x^2 + 17x +72
we would use 8 and 9, because they add to 17. The factors
would be (x+8)(x+9)

Answer 12

Oh, so the 3rd number shows you wether you will have negatives in your factors or not?

Answer 13

Yes!

Answer 14

not to butt in but in enlgish what you are looking for is two numbers whose produce is -72 and whose sum is -1
i.e. when you multiply you get -72 and when you add you get -1

just thought i would mention it

Answer 15

Do you always have to subtract to find get your second number out of the factors, or does that also depend on some other number?

Answer 16

If the third number is negative, you must subtract the pair to get the 2nd number.

Answer 17

And if its positive you add?

Answer 18

example x^2+20x - 21
factors
1 * 21
3*7
21 - 1 = 20 so that
is what we want
and the answer is (x+21)(x-1)

Answer 19

and if the 3rd number is positive you add the pair.
example x^2 +22x +21
we use the pair 1,21 because 1+21= 22 (and the 3rd number 21 is positive)
(x+1)(x+21)

Answer 20

What if the second number is negative and the third number is positive?

Answer 21

you end up with complex numbers, and have to use the quadratic formula. You may not have learned of it?

Answer 22

Is the quadratic formula the one thats really long and confusing?

Answer 23

not for me!

Answer 24

\[a ^{2}+bx+c\]
the roots are \[x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]

Answer 25

... That looks pretty long and confusing...

Answer 26

that should be
ax^2

Answer 27

practice helps

Answer 28

Haha, alright, ill practice late on those types of problems. (: Thanks for your help! I actually understand it.

Answer 29

Yep, I'd leave it for another day... no need to overload, and fry your brains!

Answer 30

Haha, i wouldnt want that to happen! (:

Answer 31

Hi, I have to correct myself. You asked
What if the second number is negative and the third number is positive?

What you do is still "add" but use negative numbers. For example
x^2 -4x +3
the factors for 3 is just the one pair 1*3, add to get 4. But the 2nd number is negative, so we should really add -1 - 3 = -4. So our two numbers are -1 and -3, and our factors
are (x-1)(x-3)
which multiples out to x^2 -x -3x +3 = x^2 -4x +3

Hope that's clear....