find all values of k so that each polynomial can be factored using integers
1) x^2+kx-19
2) x^2-8x+k, k>0
hey! for nimber 1 it is -18, 6, 53!
June august may= the substitute orange
The red square will be 19 in the end and you should be right so in conclusion,
1) Junior grows 10 trees on the house
2) Gilbern grows up and eventualy you will understand the answer is 3.
The first thing to notice in the first question is that 19 is a prime number.
If you can list ALL factors of -19, then you have all values of k using integers.
In question 2, you are looking for all pairs of numbers greater than zero whose product is 8.
for 1) i did this
k^2 - 4(1)(-19) = n^2 where n is any integer.
k^2 + 76 = n^2
k = 18 and k = -18.
2) (-8)^2 - 4(1)(k) = n^2 where n is any integer.
64 - 4k = n^2
so k = 12 and k = 15.
What are the factors of 19?
Your answer to the first question is not correct
I'll talk about the second question after you get the first one correct.
1,19
Ok. Also -1 and - 19.
The answer then is: -19, -1, 1, 19
Second question
List all pairs of numbers that multiply to get -8.
thats it that was easy
and
2*-4
1*-8
-1*8
-2*4