Answer:
The Roots of f(x) are
x=1
x=[tex]\frac{-3+\sqrt{41} }{4}[/tex]
and x=[tex]\frac{-3-\sqrt{41} }{4}[/tex]
Step-by-step explanation:
Given function f(x)=[tex]2x^{3} +x^{2} -7x+4[/tex]
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)=[tex]2x^{3} +x^{2} -7x+4[/tex]
f(1)=2+1-7+4=0
Therefore, (x-1) is one of root.
Let x=(-1)
f(1)=[tex]2x^{3} +x^{2} -7x+4[/tex]
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.
[tex]\frac{f(x)}{(x-1)} =\frac{2x^{3} +x^{2} -7x+4}{(x-1)} \\\frac{f(x)}{(x-1)} =\frac{2x^{3} +(-2x^{2}+2x^{2})+x^{2} -7x+4}{(x-1)} \\\frac{f(x)}{(x-1)} =\frac{2x^{2}(x-1)+2x^{2} + x^{2} -7x+4}{(x-1)} \\\frac{f(x)}{(x-1)} =\frac{2x^{2}(x-1)+3x^{2} -7x+4}{(x-1)}\\\frac{f(x)}{(x-1)} =\frac{2x^{2}(x-1)+3x(x-1) +3x-7x+4}{(x-1)}\\\frac{f(x)}{(x-1)} =\frac{2x^{2}(x-1)+3x(x-1)-4x+4}{(x-1)}\\\frac{f(x)}{(x-1)} =\frac{2x^{2}(x-1)+3x(x-1)-4(x-1)}{(x-1)}\\\frac{f(x)}{(x-1)} =2x^{2}+3x-4\\[/tex]
For 2x^{2}+3x-4
a=2,b=3 and c=(-4)
D=[tex]b^{2}-4ac=3^{2}-4(2)(-4)[/tex]
D=[tex]9+32[/tex]
D=[tex]41[/tex]
x=[tex]\frac{-b+\sqrt{D} }{2a}[/tex] and x=[tex]\frac{-b-\sqrt{D} }{2a}[/tex]
x=[tex]\frac{-3+\sqrt{41} }{2(2)}[/tex] and x=[tex]\frac{-3-\sqrt{41} }{2(2)}[/tex]
x=[tex]\frac{-3+\sqrt{41} }{4}[/tex] and x=[tex]\frac{-3-\sqrt{41} }{4}[/tex]
Thus,
The Roots of f(x) are
x=1
x=[tex]\frac{-3+\sqrt{41} }{4}[/tex]
and x=[tex]\frac{-3-\sqrt{41} }{4}[/tex]
