Question

show that the polynomial x^2-6x+12 has no zeroes
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Answer 1

Step-by-step explanation:

The first term is,  x2  its coefficient is  1 .

The middle term is,  -6x  its coefficient is  -6 .

The last term, "the constant", is  +12 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 12 = 12 

Step-2 : Find two factors of  12  whose sum equals the coefficient of the middle term, which is   -6 .

 -12   +   -1   =   -13     

-6   +   -2   =   -8     

-4   +   -3   =   -7     

-3   +   -4   =   -7     

-2   +   -6   =   -8     

-1   +   -12   =   -13     

1   +   12   =   13    

  2   +   6   =   8     

3   +   4   =   7    

 4   +   3   =   7     

6   +   2   =   8     

12   +   1   =   13

Observation : No two such factors can be found !! 

Conclusion : Trinomial can not be factored

Answer 2

Answer:

Step-by-step explanation:

-b+-under root b^2-4ac whole divided by 2a will give........ 6+- under root 36-48 whole divided by 2....... here we cannot notice that -12 will come in under root.... therefore it will have imaginary roots.i.e.,no real zeros