show that the polynomial x^2-6x+12 has no zeroes
Answers
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:
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