math questions.?

Question

x^4 + x^3 + x^2 -9x - 10 = 0. Solve for all possible solutions. There are three answers and one of them has an i.

show work.

Answer 1

since x has 4 power technically it has 4 roots

so let us see what are those

x^4+x^3+x^2-9x-10=0

or x^4+x^3+x^2+x-10x-10

or x^3(x+1)+x(x+1)-10(x+1)

The value of the function becomes zero when x = -1, so (x+1) is a factor

or, (x+1)(x^3+x-10)

now x^3+x-10 becomes zero when x=2, thus x-2 is a factor for this. So we get,

x^3+x-10

or, x^3-2x^2+2x^2-4x+5x-10

or, x^2(x-2)+2x(x-2)+5(x-2)

or, (x-2)(x^2+2x+5)

now the factor x^2+2x+5 gives an imaginary root and to find it we have to solve using the quadratic eualtion rule

so two roots of x^2+2x+5 would be

[- 2 +/- sqrt (-16)]/2

or [-1+/- 2i] sqrt (-1) = i

so the given euation has four roots

x1 = -1
x2 = 2
x3 = -1 + 2i
x4 = -1 - 2i

where i stands for sqrt(-1)

cool :)

Answer 2

x^4+x^3+x^2-9x-10=0
from a cursory examination we can know -1 is a root
verifying
f(-1)=1-1+1+9-10=0
so x+1 is a factor
f(2)=2^4+2^3+2^2-9(2)-10
=16+8+4-18-10=0
so x-2 is a factor
dividing by x-2 and x+1 synthetically
1 1 1 -9 -10
0 -1 0 -1 10
1 0 1 -10
so quotient on dividing by x+1 is x^3+x-10
1 0 1 -10
0 2 4 10
1 2 5 0
quotient on dividing by x+2 is x^2+2x+5
solving this by the quadratic formula
x=[-2+/-rt(4-20)]/2
=-2+/-4i /2
=-1+/-2i
so the four roots are -1,2,-1+2i,-1-2i
remember roots with i will always occur in pairs
they are called conjugate roots

Answer 3

There are 4 roots, 2 real and 2 complex
..... ..... .. x^2 + 2x + 5
x^2 - x - 2)x^4 + x^3. + x^2 -9x - 10 = 0
..... ..... . -x^4 + x^3 + 2x^2
..... ..... ..... ... 2x^3 + 3x^2 - 9x - 10
..... ..... ..... .. -2x^3 + 2x^2 + 4x
..... ..... ..... ..... ..... .. 5x^2 - 5x - 10
..... ..... ..... ..... ..... . -5x^2 + 5x + 10
(x + 1)(x - 2)(x^2 + 2x + 5) = 0
x = (-2 ± √(4 - 20))/2
x = -1 ± √(1 - 5)
x = -1 + i2, -1 - i2, -1, 2