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If f(x)=(2x-1)(x^2+1)(x-5)^2, then f(x) has how many real roots? a. 0 b. 1 c. 2 d....

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vass3 | eNoter

Posted August 5, 2013 at 5:19 PM via web

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If f(x)=(2x-1)(x^2+1)(x-5)^2, then f(x) has how many real roots?

a. 0

b. 1

c. 2

d. 3

e. 4

1 Answer | Add Yours

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 5, 2013 at 5:38 PM (Answer #1)

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Determine the number of real roots for

`f(x)=(2x-1)(x^2+1)(x-5)^2`

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.

The graph:

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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