Exercise 25 Page 322

You can find the real roots by graphing. Then, use synthetic division to find the remaining expression after factoring out the real roots.

Real roots: x=- 4, x=4
Complex roots: x=- 1+isqrt(3)/2, x=- 1-isqrt(3)/2

Practice makes perfect

Let's enter the function in the calculator by pushing Y= and writing the right-hand side of the equation in the first row.

Next, by pushing GRAPH, the calculator will draw the function.

We can see that the function has two real roots, x=- 4 and x=4. This means we will be able to factor out (x+4) and (x-4) from the equation. By using synthetic division twice, we can find the remaining expression. Let's start by factoring out (x+ 4) from the original equation.

Next, we will perform one more synthetic division on our result and factor out (x-4). Note that we can remove the remainder 0 from our calculations this time.

Let's see how these divisions look if we were to express them as polynomial factoring. This will help us visualize the coefficients given as the quotients of the synthetic division.

f(x)=x^4+x^3-15x^2-16x-16

FactorOut

Factor out (x+4)

f(x)=(x+4)(x^3-3x^2-3x-4)

FactorOut

Factor out (x-4)

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

To find the complex roots, we will use the Quadratic Formula with the remaining quadratic equation, x^2+x+1.

x=- b± sqrt(b^2-4ac)/2a

Substitute a= 1, b= 1, c= 1

x=- 1± sqrt(1^2-4( 1)( 1))/2( 1)

BaseOne

1^a=1

x=- 1± sqrt(1-4(1)(1))/2(1)

Multiply

x=- 1± sqrt(1-4)/2

SubTerm

Subtract term

x=- 1± sqrt(- 3)/2

SqrtNegToISqrt

sqrt(- a)= isqrt(a)

x=- 1± isqrt(3)/2

The complex zeros are x= - 1+isqrt(3)2 and x= - 1-isqrt(3)2. Let's summarize! Real Roots:& x=- 4, x=4 Complex Roots:& x=- 1+isqrt(3)/2, & x=- 1-isqrt(3)/2

Subchapter links

Lesson Check p.322

Practice and Problem-Solving Exercises p.322-324

Standardized Test Prep p.324

Mixed Review p.324

Exercise 25 Page 322

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