Question

Solve each equation given the indicated root:A) 2m^3 - 5m^2 -13m -5 =0; -1/2B) 2x^4 -9x^3 + 13x^2 -x -5 =0 ; 2 + i

Answer 1

Part A

Divide the given polynomial by (m+1/2)

so

2m^3 - 5m^2 -13m -5 : (m+0.5)

2m^2-6m-10

-2m^3-m^2

-----------------------------

-6m^2-13m-5

6m^2+3m

---------------------

-10m-5

10m+5

-----------

0

therefore

2m^3 - 5m^2 -13m -5=(x+0.5)(2m^2-6m-10)

Now, solve the quadratic equation

2m^2-6m-10=0

a=2

b=-6

c=-10

substitute in the formula

`m=(-(-6)\pm√(-6^2-4(2)(-10)))/(2(2))`
`m=(6\pm√(116))/(4)`

Therefore

The roots of the given equation are

`\begin{gathered} m=-(1)/(2) \\ \\ m=(6+√(116))/(4) \\ \\ m=(6-√(116))/(4) \end{gathered}`

Part B

we have

2x^4 -9x^3 + 13x^2 -x -5=0

one root is (2+i)

by the conjugate theorem

another root must be equal to (2-i)

The factors are

[x-(2+i)] and [x-(2-i)]

Multiply both factors

[x-(2+i)]*[x-(2-i)]=x^2-x(2-i)-x(2+i)+(2+i)(2-i)=x^2-2x+xi-2x-xi+4-i^2

simplify

x^2-4x+4-i^2

i^2=-1

substitute

x^2-4x+4-(-1)

x^2-4x+5

Now, divide

2x^4 -9x^3 + 13x^2 -x -5 by (x^2-4x+5)

2x^4 -9x^3 + 13x^2 -x -5 : (x^2-4x+5)

2x^2-x-1

-2x^4+8x^3-10x^2

----------------------------------

-x^3+3x^2-x-5

x^3-4x^2+5x

---------------------

-x^2+4x-5

x^2-4x+5

---------------

0

Now solve the quadratic equation 2x^2-x+1 using the formula

2x^2-x+1=0

a=2

b=-1

c=1

substitute

`x=(1\pm√(-1^2-4(2)(1)))/(4)`
`x=(1\pm i√(7))/(4)`

therefore

The roots are

`\begin{gathered} x=2+i \\ x=2-i \\ x=(1+\imaginaryI√(7))/(4) \\ \\ x=(1-\imaginaryI√(7))/(4) \end{gathered}`