determine ALL integral values of k so that each trinomial can be factored:
1) x^2 + kx - 9
2) 3x^2 + kx + 7
determine TWO integral values of k so that each trinomial can be factored:
1) x^2 - 3x + k
2) 3x^2 + 4x + k
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1) x^2 + kx - 9
The integers we are looking for here multiply to -9
These are -9 and 1 , -1 and 9 , -3 and 3
And k will sum to these, i,e. , k = -8, k = 8 and k =0
however, if k = 0, we won't have a trinomial..!!
2) 3x^2 + kx + 7
The integers that we are looking for here can be found thusly
(3x + 1) ( x + 7) → 3x^2 + 22x + 7 → k = 22
(3x + 7) ( x + 1) → 3x^2 + 10x + 7 → k = 10
1) x^2 - 3x + k
Consider that we have
(x + m) (x + n) = x^2 + (m + n)x + mn
So (m + n) = -3 and k = mn
Two integers that work are -7and 4...and their product will be k = -28
And another possibility is -9 and 6.......and their product is k = - 54
2) 3x^2 + 4x + k
To find a suitable pair, consider
(3x + m) ( x + n) = 3x^2 + (m + 3n) x + mn
So....m + 3n will add to 4 and k will be their product
This works if m = n = 1 and their product is k = 1
Another possibility is if m = `10 and = -2 and their product = k = -20
