POLYNOMIAL REVIEW

Actually...there are 7 roots.....3 real and 4 non-real

 x = ± √5   and x =  -3       are the real roots

And there are 4 non-real roots [ two conjugate pairs, as expected]

Five

Here are the roots:

x = -3
x = -sqrt(5)
x = sqrt(5)

Complex roots:
x = -2 i
x = 2 i
x = -i sqrt(10)
x = i sqrt(10)

how many roots does the following have

p(x)=(x^2-5)(x^2+4)(x^2+10)(2x+6)

$\small{ \begin{array}{rcrcrcrcrcr} && (x^2-5) &\times& (x^2+4) &\times& (x^2+10) &\times& (2x+6) \\ &=& (x^2-5) &\times& (x^2+4) &\times& (x^2+10) &\times& (2\cdot(x+3)) \\ &=& \underbrace{(x^2-5)}_{\text{binomial} } &\times& (x^2+4) &\times& (x^2+10) &\times& (x+3) &\times& 2 \\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& (x^2+4) &\times& (x^2+10) &\times& (x+3) &\times& 2 \\ && && \boxed{~ \begin{array}{rcll} i^2 = -1 \\ -4i^2 = 4 \\ \end{array} ~} && \boxed{~ \begin{array}{rcll} i^2 = -1 \\ -10i^2 = 10 \end{array} ~}\\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& (x^2-4i^2) &\times& (x^2-10i^2) &\times& (x+3) &\times& 2 \\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& \underbrace{(x^2-4i^2)}_{\text{binomial} } &\times& \underbrace{(x^2-10i^2)}_{\text{binomial} } &\times& (x+3) &\times& 2 \\ &=& (x-\sqrt{5})\cdot (x+\sqrt{5}) &\times& (x-2i)\cdot (x+2i)&\times&(x-\sqrt{10}i)\cdot (x+\sqrt{10}i)&\times& (x+3) &\times& 2 \\\\ &=& \underbrace{(x-\sqrt{5})}_{1.\text{ root: }\sqrt{5}} \cdot \underbrace{(x+\sqrt{5})}_{2.\text{ root: }-\sqrt{5}} &\times& \underbrace{(x-2i)}_{3.\text{ root: }2i} \cdot \underbrace{(x+2i)}_{4.\text{ root: }-2i} &\times& \underbrace{(x-\sqrt{10}i)}_{5.\text{ root: }\sqrt{10}i} \cdot \underbrace{(x+\sqrt{10}i)}_{6.\text{ root: }-\sqrt{10}i} &\times& \underbrace{(x+3)}_{7.\text{ root: }-3} &\times& 2 \\ \end{array} }$