If f(x)=(2x-1)(x^2+1)(x-5)^2, then f(x) has how many real roots?
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Determine the number of real roots for
f(x) has been factored into linear and irreducible quadratic factors.
The number of real roots can be found using the zero product property.
(2x-1)=0 ==> `x=1/2`
`x^2+1=0` has no real solutions.
`(x-5)^2=0==>x=5` , but this is a "double root".
f(x) is a quintic polynomial (degree 5) so there are 5 roots. The possibilities are 1real and 4 imaginary, 3 real and 2 inmaginary, or 5 real roots. We have found 3 real roots and the other factor has 2 imaginary roots.
The answer is (d) 3 real roots.
Note that the double root at x=5 appears on the graph as the point where the curve touches the x-axis and then goes back up. This is characteristic of roots to even powers.
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