1. (x^2+1)*(x^3+2*x)*(x^2-64)
=(x^2+1)*x*(x^2+2)*(x+8)(x-8)
Solving for each factor in turn, for example,
x^2+1=0 => x^2=-1 => x=+i, x=-i
x=0 => x=0
x^2+2=0 => x^2=2 => x=+sqrt(2)i, -sqrt(2)i
x+8=0 => x=-8
x-8=0 => x=+8
we have solution set
S, whereS={+i, -i, 0, +sqrt(2)i, -sqrt(2)i, -8, +8)
2. A.
x^4-81=0 => x^4=81 => x^2=+9 or x^2=-9
x^2=+9 => x=+3, -3
x^2=-9 => x=+3i, -3i
S={+3i, -3i, +3, -3}
B.
x^4+10x^2+25=0 => (x^2+5)^2=0 => ± (x^2+5)=0 => x^2=-5
=> x=+sqrt(5)i (multiplicity 2 and x=-sqrt(5)i (multiplicity 2)
S={+sqrt(5)i (multiplicity 2) -sqrt(5)i (multiplicity 2)}
C.
x^4-x^2-6=0 => (x^2-3)(x^2+2)=0 => x^2=3 or x^2=-2
S={+sqrt(2)i,-sqrt(2)i, +sqrt(3), -sqrt(3) }
3.
x^4+3x^2-4=0 = (x^2-1)(x^2+4) => x^2=1 or x^2=-4
S={+2i, -2i, +1, -1}
You need to first multiply 100 by 42 then divide by 15
Answer:
b^2√30b
Step-by-step explanation:
√30b^5 - Original
Break down the radical with its factors (factor tree)
√3*2*5*b*b*b*b*b
See what perfect squares you can take out (Can take out four b terms) You cant take out any perfect squares involving 30 since there are no perfect squares
b^2√3*2*5*b
b^2√30b
It means 1.8 × 10 to the tenth power. the E is a ten are you working with scientific notation
Answer:
the answer is 22
Step-by-step explanation:
-|-7|+|-2+17
the 7 becomes positive
(7|+|-2)+17
5+17
=22
