I have a worksheet in my hon. algebra 2 class and the question is:

State the possible number of real roots for each equation. Then find all roots.

The 1st question is:

$(x-7)(x+1)(x-1) = 0$

So would there be 3 real roots/0 imaginary roots, and the actual roots would be $7, -1,$ and $1$ since that's where it intercepts the X?

Also, another question I have, if a polynomial is squared like:

$(x-1)^2(x+7) = 0$

Would that mean the $(x-1)^2$ would be an imaginary root which would make the answer 1 real root/1 imaginary root and then $1, -7$ as the roots?

Thanks a lot!

  • $\begingroup$ No, it has a double root. Also, complex roots will always come in complex conjugate pairs. For example $x^2 + 1 = 0 \implies x_{1,2} = \pm i$. That is, we can factor it as $(x+i)(x-i) = 0$. $\endgroup$
    – Moo
    Jan 22, 2017 at 22:28
  • $\begingroup$ No. In the context of roots of single variable polynomials roots are counted more than once so that the number of roots equals the degree of the polynomial. So the "three" roots of $(x-1)^2(x+7)$ are $1,1,-7$ because the three factors are $(x-1)(x-1)(x+7)$--one root for each factor, even if it is repeated. $\endgroup$ Jan 22, 2017 at 22:29

1 Answer 1


If a polynomial has a factor such as $(x-a)^n$ it is named as multiplicity, not an imaginary root. Imaginary root is when delta<0.

For example let $(x^2+1)(x-2)^2=0$

Here you have imaginary roots $i$ and $-i$ from $(x^2+1)$ and double roots $2$ and $2$ from $(x-2)^2$.


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