According to the Rational Root Theorem, integer solutions of this equation are factors of the constant term 8.
Let's list all possibilities.
± 1, ± 2, ± 4,± 8
We can use synthetic division to try these possibilities.
Let's start with x=1.
rl IR-0.15cm r 1 & |rr 1& -5& 3& 2& 8 & 1& -4& -1& 1 & c 1& -4& -1 & 1& 9
Since the last entry of the last line is 9, x=1 is not a solution.
Let's try x=-1 next.
rl IR-0.15cm r -1 & |rr 1& -5& 3& 2 & 8 & -1& 6& -9 & 7 & c 1& -6& 9& -7& 15
Since the last entry of the last line is 15, x=-1 is not a solution.
Let's try x=2 next.
rl IR-0.15cm r 2 & |rr 1& -5& 3& 2& 8 & 2& -6& -6& -8 & c 1& -3& -3& -4 & 0
Since the last entry of the last line is 0 we know that x=2 is a solution, and we also get the factor form of the expression.
x^4-5x^3+3x^2+2x+8=(x-2)(x^3-3x^2-3x-4)
To find other solutions, we repeat this process.
This time it is enough to work with the cubic factor.
Since the constant term in this factor is -4, the possible integer roots are now the factors of -4.
± 1, ± 2, ± 4
We have already found that ± 1 are not solutions, so we will only check the other possibilities.
Since x=2 might be a double zero of the original equation, we will try it again.
rl IR-0.15cm r 2 & |rr 1& -3& -3& -4 & 2& -2& -10 & c 1& -1& -5& -14
Since the last entry of the last line is -14, x=2 is not a root of the cubic factor.
Next we try x=-2.
rl IR-0.15cm r -2 & |rr 1& -3& -3& -4 & -2& 10& -14 & c 1& -5 & 7 & -18
Since the last entry of the last line is -18, x=-2 is not a root of the cubic factor.
Next we try x=4.
rl IR-0.15cm r 4 & |rr 1& -3& -3& -4 & 4& 4& 4 & c 1 & 1 & 1 & 0
Since the last entry of the last line is 0, we know that x=4 is a solution. We also get the factor form of the expression.
x^4-5x^3+3x^2+2x+8&=(x-2)(x^3-3x^2-3x-4)
&=(x-2)(x-4)(x^2+x+1)
To find the other solutions, we can set the quadratic factor equal to 0 and solve the equation.
x^2+x+1=0
Let's check the discriminant first.
b^2-4ac=1^2-4(1)(1)=-3
This is negative, so there are no real solutions.
The only solutions of f(x)=g(x) are the ones we found above using synthetic division.
Solutions: x=2,x=4
Graphing Approach
Let's use a graphing calculator to check our answer.
We begin by pushing the Y= button and typing the equations in the first two rows.
To see the graphs, you may need to change the viewing window.
Push WINDOW, change the settings, and push GRAPH.
Next, to find the intersection, push 2nd and TRACE. From this menu, choose intersect.
The calculator will prompt you to choose the first and second curve, and to provide the calculator with a best guess of where the intersection might be.
You will need to repeat this process twice to find both solutions.
We can see that the algebraic and graphing approach gave the same solutions.
Solutions: x=2,x=4
