Exercise 186 Page 434

We want to find the roots of the polynomial function f(x). To do so, we need to solve the equation f(x)=0. x^3+3x^2+x-5 = 0 The degree of f(x) is 3. Thus, by the Fundamental Theorem of Algebra we know that the equation has exactly three solutions. To find them we will perform the following steps.

1. First, use the graph to find the first real root of the function.
2. Then, use polynomial long division to factor out the binomial corresponding to the first root in the equation.
3. Finally, solve the remaining quadratic equation using the Quadratic Formula.

Using the Graph

Note that (1,0) is the point where the graph of the function intercepts the x-axis. Therefore, x=1 is the first root of the function and the solution to the equation.

Dividing the Polynomial

We have already found that x=1 is a solution to f(x)=0. Therefore, (x-1) is a factor of the polynomial function. We will now divide the polynomial by (x-1) to find the remaining roots. (x^3+3x^2+x-5)÷ (x-1) To divide the polynomial we can use polynomial long division. When doing that, all the terms of the dividend must be present and the polynomial must be in standard form. x^3+3x^2+x-5 Since there are no missing terms and our polynomial is in descending degree order, we do not need to rewrite the polynomial. Let's divide the given polynomial by (x-1).

l r x-1 & |l x^3+3x^2+x-5

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Divide

x^3/x= x^2

r x^2 r x-1 & |l x^3+3x^2+x-5

Multiply term by divisor

r x^2 rl x-1 & |l x^3+3x^2+x-5 & x^3-x^2

Subtract down

r x^2 r x-1 & |l 4x^2+x-5

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Divide

4x^2/x= 4x

r x^2 + 4x r x-1 & |l 4x^2+x-5

Multiply term by divisor

rx^2 + 4x rl x-1 & |l 4x^2+x-5 & 4x^2-4x

Subtract down

r x^2+4x r x-1 & |l 5x-5

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Divide

5x/x= 5

r x^2+4x+5 r x-1 & |l 5x-5

Multiply term by divisor

rx^2+4x+ 5 rl x-1 & |l 5x-5 & 5x-5

Subtract down

r x^2+4x+5 r x-1 & |l 0

Since the quotient is x^2+4x+5, we can now factor out (x-1) and obtain the following equation. (x-1)(x^2+4x+5) = 0

Factoring the Remaining Quadratic Factor

Note that as a second factor we obtained a quadratic function. We can use the Quadratic Formula to find its roots. To do so we need to identify the values of a, b, and c. x^2+4x+5=0 ⇕ 1x^2+ 4x+ 5=0 We can see above that a= 1, b= 4, and c= 5. Now, let's substitute these values into the Quadratic Formula to find the remaining roots.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 4±sqrt(4^2-4( 1)( 5))/2( 1)

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Simplify right-hand side

CalcPow

Calculate power

x=-4±sqrt(16-4(1)(5))/2(1)

Multiply

Multiply

x=-4±sqrt(16-20)/2

SubTerm

Subtract term

x=-4±sqrt(-4)/2

SqrtNegToSqrtI

sqrt(- a)= sqrt(a)* i

x = -4± sqrt(4)* i/2

CalcRoot

Calculate root

x = -4± 2i/2

We can simplify this result into two separate roots.

x=-4± 2i/2
x_1=-4+2i/2	x_2=-4-2i/2
x_1= -2+i	x_2= -2-i

We already knew that 1 was a root, but now we have found the remaining two roots. Notice that both -2+i and -2-i are complex numbers. Finally, let's list all of the roots we have found for the given function. Real Roots:& 1 Complex Roots:& -2± i