Question

Use the remainder theorem to determine which number is a root of f(x) = 2x3 + x2 - 7x + 4.

Answer 1

The Roots of f(x) are

x=1

x=
`(-3+√(41) )/(4)`

and x=
`(-3-√(41) )/(4)`

Explanation:

Given function f(x)=
`2x^(3) +x^(2) -7x+4`

We know that

A. The sum of coefficient of polynomial is zero then, (x-1) is one of root of polynomial

f(1)=0

B. The difference of sum of coefficient of odd term and sum of coefficient of even term of polynomial is zero then, (x+1) is one of root of polynomial

f(-1)=0

Let x=1

f(1)=
`2x^(3) +x^(2) -7x+4`

f(1)=2+1-7+4=0

Therefore, (x-1) is one of root.

Let x=(-1)

f(1)=
`2x^(3) +x^(2) -7x+4`

f(1)=-2+1+7+4=10

Therefore, (x-1) is not one of root.

Using Remainder theorem:

Dividing (x-1) on both the side.

`(f(x))/((x-1)) =(2x^(3) +x^(2) -7x+4)/((x-1)) \\(f(x))/((x-1)) =(2x^(3) +(-2x^(2)+2x^(2))+x^(2) -7x+4)/((x-1)) \\(f(x))/((x-1)) =(2x^(2)(x-1)+2x^(2) + x^(2) -7x+4)/((x-1)) \\(f(x))/((x-1)) =(2x^(2)(x-1)+3x^(2) -7x+4)/((x-1))\\(f(x))/((x-1)) =(2x^(2)(x-1)+3x(x-1) +3x-7x+4)/((x-1))\\(f(x))/((x-1)) =(2x^(2)(x-1)+3x(x-1)-4x+4)/((x-1))\\(f(x))/((x-1)) =(2x^(2)(x-1)+3x(x-1)-4(x-1))/((x-1))\\(f(x))/((x-1)) =2x^(2)+3x-4\\`

For 2x^{2}+3x-4

a=2,b=3 and c=(-4)

D=
`b^(2)-4ac=3^(2)-4(2)(-4)`

D=
`9+32`

D=
`41`

x=
`(-b+√(D) )/(2a)` and x=
`(-b-√(D) )/(2a)`

x=
`(-3+√(41) )/(2(2))` and x=
`(-3-√(41) )/(2(2))`

x=
`(-3+√(41) )/(4)` and x=
`(-3-√(41) )/(4)`

Thus,

The Roots of f(x) are

x=1

x=
`(-3+√(41) )/(4)`

and x=
`(-3-√(41) )/(4)`