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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

Guest Jul 21, 2017
 #1
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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 

 

 

cool cool cool

CPhill  Jul 21, 2017

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