Exercise 66 Page 351

We want to find the zeros of the polynomial function y=P(x). To do so, we need to solve the equation P(x)=0. 2x^3-3x^2+3x-2=0

The degree of P(x) is 3. Thus, by the Fundamental Theorem of Algebra, we know that P(x)=0 has exactly three roots. Let's find them.

Integer roots

By the Rational Root Theorem, we know that integer roots must be factors of the constant term. Since the constant term of P(x) is - 2, the possible integer roots are ± 1 and ± 2. Let's check.

x	2x^3-3x^2+3x-2	P(x)=2x^3-3x^2+3x-2
1	2( 1)^3-3( 1)^2+3( 1)-2	0 ✓
- 1	2( - 1)^3-3( - 1)^2+3( - 1)-2	- 10 *
2	2( 2)^3-3( 2)^2+3( 2)-2	8 *
- 2	2( - 2)^3-3( - 2)^2+3( - 2)-2	- 36 *

We found that 1 is a root for P(x)=0. Therefore, (x-1) is a factor of the polynomial. Let's use synthetic division to factor out (x-1) to find the other two roots.

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 &- 2

Bring down the first coefficient

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 &- 2 & & & & c 2& & &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 & - 2 & 2 & & & c 2& & &

Add down

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 & - 2 & 2 & & & c 2& - 1& &

▼

Repeat the process for all the coefficients

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 & - 2 & 2 &- 1 & & c 2& - 1& &

Add down

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 & - 2 & 2 &- 1& & c 2& - 1 &2 &

Multiply the coefficient by the divisor

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 & - 2 & 2 &- 1 & 2 & c 2& - 1 & 2 &

Add down

rl IR-0.15cm r 1 & |rr 2& - 3 & 3 & - 2 & 2 &- 1 & 2 & c 2& - 1 & 2 & 0

Using synthetic division to remove the first root, 1, left us with the following polynomial. 2x^2-x+2

Factoring the remaining quadratic factor

We will use the Quadratic Formula to find the remaining factors. To do so, we will need to identify the values of a, b, and c. 2x^2-x+2 ⇔ 2x^2 -1x+ 2=0 We can see above that a= 2, b= - 1, and c= 2. Finally, we will substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -1)±sqrt(( -1)^2-4( 2)( 2))/2( 2)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=1±sqrt((-1)^2-4(2)(2))/2(2)

CalcPow

Calculate power

x=1±sqrt(1-4(2)(2))/2(2)

Multiply

Multiply

x=1±sqrt(1-16)/4

SubTerm

Subtract term

x=1±sqrt(- 15)/4

SplitIntoFactors

Split into factors

x=1±sqrt(- 1* 15)/4

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=1 ± sqrt(- 1) sqrt(15)/4

SqrtNegOneToI

sqrt(- 1)= i

x=1 ± isqrt(15)/4

We have found our other two roots, 1+isqrt(15)4 and 1-isqrt(15)4. Finally, let's list all of the roots we have found for the given function.

x=1, 1+isqrt(15)/4,and 1-isqrt(15)/4