Is X^2 - 16x + 63 prime? What is the factored version?
what is x equal to ???
I don't know
Set up the parenthesis ( ) ( ) Distribute x^2 (x ) (x ) Now you need to find factors of 63 that will add to be -16. Can you name the factors of 63 for me please?
umm, 1, 3, 7, 9, 21, 63?
Okay. Good don't forget the negative factors as well!! -1, -3,- 7, -9, -21, -63
(x -21 )(x-3 )
I am not sure whether that is the correct answer
does that mean it is prime?
Yes it is prime.
No it isn't
not actually. from those factors, pick up 2 numbers whose sum is -16 and product = 63
so its x^2 - 16x +63 = 63?
what was that ? 2 numbers from -1, -3,- 7, -9, -21, -63 that add up to -16 are ... ?
-7, and -9
( x-7)(x-9) \
I figured it out all of a sudden
Sorry, my wifi cut off..good to see the others stepped in to help..
good, so you split -16x as -7x-9x x^2 - 16x +63 = x^2 - 9x-7x +63 factor x from 1st 2 terms and -7 from last two terms, what u get ?
I figured it out all of a sudden in my mind when I was doing something around the house
( -7)(-9)=63 -7-9=-16
now i'm totally lost. Is -16 the factored value? x = -16?
x shouldn't equal anything at the moment, you just needed to factor.
(x-7)(x-9) FOIL First Outer Inner Last Checking stage x^2-9x-7x+63 x^2-16x+63
x^2 is x squared
so it is prime?
It is prime because it can be factored
thanks!
It is NOT prime because it can be factored
I meant not prime
I typed in prime
so what does x^2 - 16x +63 equal?
you split -16x as -7x-9x x^2 - 16x +63 = x^2 - 9x-7x +63 factor x from 1st 2 terms and -7 from last two terms, what u get ?
If you let \(x=1\) doesn't it end up being an even number?
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