Given polynomial p(x)=x2-6x+9
list of values of p(x):
| x | -1 | 0 | 1 | 2 | 3 | 4 |
| x2 | 0 | 1 | 4 | 16 | ||
| 6x | 0 | 6 | 12 | 24 | ||
| x2-6x | 7 | 0 | -8 | -9 | ||
| 9 | 9 | 9 | 9 | 9 | 9 | 9 |
| p(x) | 9 | 1 | 0 | |||
| (x,y) | ( , ) | ( , ) | ( , ) | ( , ) | ( , ) | ( , ) |
Now,lets locate the points listed above on a graph paper
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Result:We observe that the graph cuts the X-axis at(3,0).
So,the zeros of the given polynomial are same i.e,3
Justification:
Given p(x)=x2-6x+9
x2-3x-x+9=0
x(x-3)-(x-)=0
x-3=0 and x-=0
x=3 and x=