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)
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]
how many roots does the following have
p(x)=(x^2-5)(x^2+4)(x^2+10)(2x+6)
(x2−5)×(x2+4)×(x2+10)×(2x+6)=(x2−5)×(x2+4)×(x2+10)×(2⋅(x+3))=(x2−5)⏟binomial×(x2+4)×(x2+10)×(x+3)×2=(x−√5)⋅(x+√5)×(x2+4)×(x2+10)×(x+3)×2 i2=−1−4i2=4 i2=−1−10i2=10 =(x−√5)⋅(x+√5)×(x2−4i2)×(x2−10i2)×(x+3)×2=(x−√5)⋅(x+√5)×(x2−4i2)⏟binomial×(x2−10i2)⏟binomial×(x+3)×2=(x−√5)⋅(x+√5)×(x−2i)⋅(x+2i)×(x−√10i)⋅(x+√10i)×(x+3)×2=(x−√5)⏟1. root: √5⋅(x+√5)⏟2. root: −√5×(x−2i)⏟3. root: 2i⋅(x+2i)⏟4. root: −2i×(x−√10i)⏟5. root: √10i⋅(x+√10i)⏟6. root: −√10i×(x+3)⏟7. root: −3×2